Saved Bookmarks
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
The length of the latusretum of the parabola `169|(x-1)^(2)+(y-3)^(2)|=(5x-12y+7)^(2)`, isA. `14/13`B. `28/13`C. `12/13`D. `48/13` |
|
Answer» Correct Answer - D The equation of the parabola is `sqrt((x-1)^(2)+(y-3)^(2))=|(5x-12y+7)/(sqrt(5^(2)+(-12)^(2)))|` Clearly, its focus is (1, 3) and the equation of the directrix is 5x-12y=0. `:." Length of the latusrectum " = 2 xx" Length of the perpendicualr from the focus (1, 3) on the directrix"` `rArr" Length of the latusrectum = 2"|(5xx1-12xx3+7)/(sqrt(5^(2)+(-12)^(2)))|=48/13` |
|
| 202. |
A circle is drawn through the point of intersection of the parabola `y=x^(2)-5x+4` and the x-axis such that origin lies outside it. The length of a tangent to the circle from the origin is ________ . |
|
Answer» Correct Answer - 2 (2) Point of intersection of given with x-aixs are (1,0) and (4,0). Now, circle passing through these points is `(x-1)(x-4)+y^(2)+lamday=0` `:." Length of the tangent from "(0,0)=sqrt(4)=2` |
|
| 203. |
If the length of the latus rectum rectum of the parabola `169{(x-1)^(2)+(y-3)^(2)}=(5x-12y+17)^(2)` is L then the value of 13L/4 is _________. |
|
Answer» Correct Answer - 7 (7) Here, `(x-1)^(2)+(y-3)^(2)={(5x-12+17)/(sqrt(5^(2))+(-12)^(2))}^(2)` Therefore, the is (1,3) and directrix is 5x-12y+17=0. The distance of the focus from the directrix is `|(5xx1-12xx3+17)/(sqrt(5^(2)+(-12)^(2)))|=(14)/(13)` `:." Length of latus rectum"=2xx(14)/(13)=(28)/(13)` |
|
| 204. |
The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0`, externally and also touch thex-axis, lieon :(1) a circle.(2) an ellipse which is not acircle.(3) a hyperbola.(4) a parabola. |
|
Answer» `x^2 + y^2 - 8x - 8y - 4= 0` `x^2 - 8x + 16 + y^2 - 8y+16 - 4 - 16-16= 0` `(x-4)^2 + (y- 4)^2 - 36 = 0` `(x-4)^2 + (y-4)^2 = 6^2` radius`= 6` centre= `(4,4)` centre of circle which touches the given circle externally is `sqrt((h-4)^2 + (k-4)^2) = 6 + r` `sqrt((h-4)^2 + (k-4)^2) = (6+k)` `(h-4)^2 + (k-4)^2 = (6 + k)^2` `h^2 + 16 - 8h + k^2 + 16 - 8k = 36 + k^2 + 12k` `h^2 -8h-20k + 32 - 36= 0` `h^2 - 8h - 20k - 4= 0` keeping h= x & k=y `x^2 - 8x - 20y - 4=0` this eqn is of parabola so, option 3 is correct |
|
| 205. |
Find the length of normal chord which subtends an angle of `90^0`at the vertex of the parabola `y^2=4xdot`A. `6sqrt3`B. `3sqrt3`C. `2`D. `1` |
| Answer» Correct Answer - A | |
| 206. |
If a chord, which is not a tangent, of the parabola `y^(2)=16x` has the equation 2x+y=p, and midpoint (h,k), then which of the following is (are) possible value(s) of p,h and k ?A. p=5, h=4, k=-3B. p=-1, h=1, k=-3C. p=-2, h=2, k=-4D. p=2, h=3, k=-4 |
|
Answer» Correct Answer - D 4 Parabola, `y^(2)=16x` Equation of chord, 2x+y=p (1) Equation of chord having mid point (h,k) is `T=S_(1)` `ky=8(x+h)=k^(2)-16h` `orky-8x=k^(2)=8h` comparing (1) and (2), we get `(k)/(1)=(-8)/(2)=(k^(2)-8h)/(p)` `rArrk=-4andk^(2)-8h=-4p` From this, we get 16-8h=-4p Clearly, h=3 and p=2. |
|
| 207. |
The radius of the circle whose centre is (-4,0) and which cuts the parabola `y^(2)=8x` at A and B such that the common chord AB subtends a right angle at the vertex of the parabola is equal toA. 4B. 3C. `sqrt(18)`D. 5 |
|
Answer» Correct Answer - A Let be the radius of the circle. Then, its equation is `(x+)^(2)+y^(2)=r^(2)" ...(i)"` This cute the parabola `y^(2)=8x` at points `A(x_(1), y_(1))` and `B(x_(2), y_(2))`. The abcissae of A and B are the roots of the equation `(x+4)^(2)+8x=r^(2)or, x^(2)+16x+16-r^(2)=0` `:." "x_(1)x_(2)=16-r^(2)" ....(ii)"` The ordinates of A and B are given by `y_(1)^(2)=8x_(1)" and "y_(2)^(2)=8x_(2)` respectively. `:." "y_(1)=+-2sqrt2x_(1)" and "y_(2)=+-2sqrt2x_(2)` Since AB subtends a right angle at the vertex of the parabola. `:." "y_(1)/x_(1)xxy_(2)/x_(2)=-1` `rArr" "x_(1)x_(2)+y_(1)y_(2)=0` `rArr" :x_(1)x_(2)+8x_(1)x_(2)=0` `rArr" "x_(1)x_(2)=0" "["Using (i)"]` `rArr" "16-r^(2)=0rArrr=4` |
|
| 208. |
The tangent PT and the normal PN to the parabola `y^2=4ax` at a point P on it meet its axis at points T and N, respectively. The locus of the centroid of the triangle PTN is a parabola whose:A. `"vertex is "(2a//3, 0)`B. `"Directri is x = 0"`C. `"Latusrectum is "(2a)/3`D. `"Focus is "(-a, 0)` |
|
Answer» Correct Answer - A Let `P(at^(2), 2at)` be any point on the parabola `y^(2)=4ax`. The equations of tangent and normal at P are `ty=x+at^(2)" ...(i)"` `"and, "y+tx=2at+at^(2)" ...(ii)"` These two meet x-axis i.e. axis of the parabola at `T(-at^(2), 0)` and `N(2a+at^(2), 0)` respectively. Let (h, k) be the coordinates of the centroid of `DeltaPTN`. Then, `h=(2a+at^(2))/3" and "k=(2at)/3` `rArr" "3h=2a+a((3k)/(2a))^(2)" [On eliminating t]"` `rArr" "12ah=8a^(2)+9k^(2)` Hence, the locus of (h, k) is `12ax=8a^(2)+9y^(2)` `rArr" "y^(2)=(4a)/3(x-(2a)/3)` Clearly, it represents a parabola whose vertex is at (2a/3, 0). Latusrectum `=(4a)/3`, focus `((2a)/3+a/3,0)-(a, 0)` and directrix `x=a//3`. |
|
| 209. |
Which of the following line can be tangent to the parabola `y^2=8x ?``x-y+2=0`(b) `9x-3y+2=0``x+2y+8=0`(d) `x+3y+12=0`A. x-y+2=0B. 9x-3y+2=0C. x+2y+8=0D. x+3y+12=0 |
|
Answer» Correct Answer - A::B::C 1,2,3 The equation of tangent `y^(2)=8x` having slope m is `y=mx+(2)/(m)` Options (1), (2), and (3) are tangents for m=1,3, and -1/2, respectively. |
|
| 210. |
The focal chord to `y^2=16x` is tangent to `(x-6)^2+ y^2 =2` then the possible values of the slope of this chordA. (-1, 1)B. (-2, 2)C. (-2, 1/2)D. (2, -1/2) |
|
Answer» Correct Answer - A The coordinates of the focus of the parabola `y^(2)=16x` are (4, 0). The equation of tangents of slope m to the circlen `(x-6)^(2)+(y-0)^(2)=(sqrt2)^(2)" are "y=m(x-6)+-sqrt2sqrt(1+m^(2))` If these tangents pass through the focus i.e. (4, 0) then `0=-2m+-sqrt2sqrt(1+m^(2))` `rArr" "4m^(2)=2(1+m)^(2)rArr2m^(2)=2rArrm=+-1` |
|
| 211. |
Two mutually perpendicular tangents of the parabola `y^(2)=4ax` meet the axis at `P_(1)andP_(2)`. If S is the focal of the parabola, Then `(1)/(SP_(1))+(1)/(SP_(2))` is equal toA. `(1)/(2a)`B. `(1)/(a)`C. `(2)/(a)`D. `(4)/(a)` |
|
Answer» Correct Answer - B (2) Tangents at `A(t_(1))andA(t_(2))` are respectively, `t_(1)y=x+at_(1)^(2)` `t_(2)y=x+at_(2)^(2)` Tangent meet axis at `P_(1)(-at_(1)^(2),0)andP_(2)(-at_(2)^(2),0)`. `:." "SP_(1)=a+at_(1)^(2),andSP_(2)=a+at_(2)^(2)` Since tangents are perpendicular, `t_(1)t_(2)=-1`. `:.(1)/(SP_(1))+(1)/(SP_(2))=(1)/(a+at_(1)^(2))+(1)/(a+at_(2)^(2))` `=(1)/(a+at_(1)^(2))+(1)/a+(a)/(t_(1)^(2))` `=(1)/(a+at_(1)^(2))+(t_(1)^(2))/(at_(1)^(2)+a)` `(1)/(a)` y-x-0 |
|
| 212. |
The circle x2 + y2 – 2x – 6y + 2 = 0 intersects the parabola y2 = 8x orthogonally at the point P. The equation of the tangent to the parabola at P can be(a) 2x – y + 1 = 0(b) 2x + y – 2 = 0(c) x + y – 4 = 0(d) x – y – 4 = 0 |
|
Answer» Correct option (a) 2x –y+1 = 0 Explanation: Let y = mx + 2/m be tangent to y2 = 8x. Since circle intersects the parabola orthogonally. So this tangent is the normal for the circle. Every normal of the circle passes through its centre. So centre (1, 3). 3 - m + 2/m2 - 3m + 2 = 0 (m - 2)(m -1) = 0 m = 1, 2 y = x + 2 or y = 2x + 1
|
|
| 213. |
Find the equation of the parabola with vertex at the origin, passing through the point P(5, 2) and symmetric with respect to the y-axis. |
|
Answer» The equation of a parabola with vertex at the origin and symmetric about the y-axis: x2 = 4ay As we are given, parabola is passing through the point P(5,2). Putting x = 5 and y = 2 in x2 = 4ay => 25 = 4a(2) = 8a => a = 25/8 Therefore, equation of parabola: x2 = 4(25/8 )y = 25/2 y or 2x2 = 25y |
|
| 214. |
The circle `x^2 + y^2 - 2x - 6y+2=0` intersects the parabola `y^2 = 8x` orthogonally at the point `P`. The equation of the tangent to the parabola at `P` can beA. x-y-4=0B. 2x+y-2=0C. x+y-4=0D. 2x-y+1=0 |
|
Answer» Correct Answer - D Let `y=mx+2/m" be a tangent to "y^(2)=8x.` Since the circle `x^(2)+y^(2)-2x -6y+2=0` interscts the parabola `y^(2) = 8x` through the centre of the circle i.e. (1, 3). `:." "3=m+2/mrArrm^(2)-3m+2=0rArrm=1,2` Thus, the equations of tangents are `y=x+2" and "2x-y+1=0.` |
|
| 215. |
Mutually perpendicular tangents `T Aa n dT B`are drawn to `y^2=4a x`. Then find the minimum length of `A Bdot`A. aB. 2aC. 4aD. 8a |
|
Answer» Correct Answer - C We know that tangents at the ends of focal chord are perpendicualar. Thus, the chord of contanct of perpendicualar tangents is a focal chord. Also, latusrectum is a focal chord of minimum length. Hence, the minimum length of AB is 4a. |
|
| 216. |
Let PQ be a focal chord of the parabola `y^2 = 4ax` The tangents to the parabola at P and Q meet at a point lying on the line `y = 2x + a, a > 0`. Length of chord PQ isA. 7aB. 5aC. 2aD. 3a |
|
Answer» Correct Answer - B Let `(at_(1)^(2),2at_(1))" and "Q(at_(2)^(2),at_(2))` be the end-point of a focal chord of the parabola `y^(2)=4ax`. Then, `t_(1)t_(2)=-1" and "PQ=a(t_(2)-t_(1))^(2)`. The tangents at P and Q intersect at a point `(at_(1)t_(2),a(t_(1)+t_(2)))` which lies on y = 2x + a. `:." "e(t_(1)+t_(2))=2at_(1)t_(2)+a` `rArr" "t_(1)+t_(2)=-2+1+1` `rArr" "t_(1)+t_(2)=-2+1=-1" "[becauset_(1)t_(2)=-1]` `:." "PQ=a(t_(2)-t_(1))^(2)` `=a{(t_(2)+ty_(1))^(2)-4t_(1)t_(2)}=a[(-1)^(2)-4(-1)]=5a` |
|
| 217. |
A variable circle passes through the fixed point (2, 0) and touches y-axis Then, the locus of its centre, isA. a parabolaB. a circleC. an ellipseD. a hyperbola |
| Answer» Correct Answer - A | |
| 218. |
A circle touches the `x`-axis and also touches the circle with center `(0, 3)` and radius `2`. The locus of the centerA. a circleB. an ellipseC. a parabolaD. a hyperbola |
|
Answer» Correct Answer - C (3) Let `C_(1)(h,k)` be the centre of the circle. The circle touches the c-axis. Then its radius is `r_(1)=k` Also, the circle touches the circle with center `C_(2)(0,3)` and radius `r_(2)=2`. Therefore, `|C_(1)C_(2)|=r_(1)+r_(2)` `orsqrt((h-0)^(2)+(k-3)^(2))=|k+2|` Squaring, we get `h^(2)-10k+5=0` Therefore, the locus is `x^(2)-10y+5=0`, which is a parabola. |
|
| 219. |
If `t` is the parameter for one end of a focal chord of the parabola `y^2 =4ax,` then its length is :A. `a(t+1/t)^(2)`B. `a(t-1/t)^(2)`C. `a(t+1/t)`D. `a(t-1/t)` |
| Answer» Correct Answer - A | |
| 220. |
The Cartesian equation of the directrix of the parabola whose parametrix equations are `x=2t+1, y=t^(2)+2`, isA. y = 2B. y = 1C. y = -1D. y = -2 |
| Answer» Correct Answer - B | |
| 221. |
Show that the curve whose parametric coordinates are `x=t^(2)+t+l,y=t^(2)-t+1` represents a parabola. |
|
Answer» From the given relations, we have `(x+y)/(2)=t^(2)+1,(x-y)/(2)=t` Eliminating t, we get `2(x+y)=(x-y)^(2)+4` This is second-degree equation in which second-degree terms form perfect square. Rewriting the equation, we have `x^(2)+y^(2)-2xy-2x-2y+4=0` Comparing with `ax^(2)+by^(2)+2hxy+2gx+2fy+c=0`, we find that `abc+2fgh-af^(2)-bg^(2)-ch^(2)!=0`. So, given equation represents a parabola. |
|
| 222. |
The parametric equation of a parabola is `x=t^2+1,y=2t+1.`Then find the equation of the directrix. |
|
Answer» We have `x=t^(2)+1,y=2t+1`. Eliminating t, we get `x-1=((y-1)/(2))^(2)` `or" "(y-1)^(2)=4(x-1)` Comparing with `(y-k)^(2)=4a(x-h)`, we have vertex (1,1) and axis parallel to x-axis having equation y-1=0. Also, 4a=4. So, a = 1 Directrix is parallel to y-axis and lies to the left of the vertex at distance a units from it. `So, equation of directrix is x=1-1 or x=0. |
|
| 223. |
A point `P(x , y)`moves in the xy-plane such that `x=acos^2theta`and `y=2asintheta,`where `theta`is a parameter. The locus of the point `P`is a/ancircle(b) aellipseunbounded parabola (d) part of the parabolaA. circleB. ellipseC. unbounded parabolaD. part of the parabola |
|
Answer» Correct Answer - D (4) Eliminating `theta` from the given equations, we get `y^(2)=-4a(x-a)` which is a parabola but `0lecos^(2)thetale1` `or" "0lexlea` `and" "-1lesinthetale1` `or" "-2aleyle2a` Hence, the locus of point P is not whole parabola, but it is a part of the parabola. |
|
| 224. |
If the chord joining the points `t_1`and `t_2` on the parabola `y^2 = 4ax` subtends a right angle at its vertex then `t_2=` |
|
Answer» Correct Answer - B The equation of the chord joining points `(at_(1)^(2), 2at_(1))"and"(at_(2)^(2), 2at_(2))` is `y(t_(1)+t_(2))=2x+2at_(1)t_(2)" ...(i)"` It is given that the axis i.e. y = 0, directrix x= -a and (i0 are concurrent. Therefore, chord (i) passes through the point of intersection of axis y = 0 adn directrix x= -a i.e. the point (-a, 0). `:." "-2a+2at_(1)t_(2)=0rArrt_(1)t_(2)=1` |
|
| 225. |
From a fixed point A three normals are drawn to the parabola `y^(2)=4ax` at the points P, Q and R. Two circles `C_(1)" and "C_(2)` are drawn on AP and AQ as diameter. If slope of the common chord of the circles `C_(1)" and "C_(2)` be `m_(1)` and the slope of the tangent to teh parabola at R be `m_(2)`, then `m_(1)xxm_(2)`, is equal toA. `1/2`B. `2`C. `-1/2`D. `-2` |
| Answer» Correct Answer - A | |
| 226. |
Find the equations of the normals at the ends of the latus- rectum of the parabola `y^2= 4ax.` Also prove that they are at right angles on the axis of the parabola.A. `x^(2)-y^(2)-6ax+9a^(2)=0`B. `x^(2)-y^(2)-6ax-6an+9a^(2)=0`C. `x^(2)-y^(2)-6xy+9a^(2)=0`D. none of these |
| Answer» Correct Answer - A | |
| 227. |
The tangents to the parabola `y^2=4a x`at the vertex `V`and any point `P`meet at `Q`. If `S`is the focus, then prove that `S PdotS Q ,`and `S V`are in GP. |
|
Answer» Let the parabola be `y^(2)=4ax`. Q is the intersection of the line x=0 and the tangent at point `P(at^(2),2at),ty=x+at^(2)`. Solving these, we get Q = (0, at). Also, S=(a,0). Now, focal length `SP=a+at^(2)` `SQ^(2)=a^(2)+a^(2)t^(2)=a^(2)(t^(2)+1)` and SV = a `:." "SQ^(2)=SPxxSV` Therefore, SP,SQ and SV are in GP. |
|
| 228. |
The tangents to the parabola `y^2=4a x`at the vertex `V`and any point `P`meet at `Q`. If `S`is the focus, then prove that `S PdotS Q ,`and `S V`are in GP.A. `A.P.`B. `G.P.`C. `H.P.`D. none of these |
|
Answer» Correct Answer - B Let `y^(2)=4ax` be a parabola with vertex at A(0, 0) and `P(at^(2), 2at)` be any point on it. The equation of tangents at A and P are x=0 and `ty=x+at^(3)` respectively. These two intersect at Q(0, at). The focus S of dsthe parabola `y^(2)=4ax` has coordinates (a, 0). `:." "SP=a+at^(2),SQ=ssqrt(1+t^(2))" and "SA=a` ltrbgt `rArr" "SQ^(2)=SPxxSArArrSP, SQ, SA" are in G.P."` |
|
| 229. |
Which one of the following equation represent parametric equation to aparabolic curve?`x=3cost ; y=4sint``x^2-2=2cost ; y=4cos^2t/2``sqrt(x)=tant ;sqrt(y)=sect``x=sqrt(1-sint ;)y=sint/2+cost/2`A. `x=3cost,y=4sint`B. `x^(2)-2=2cost,y=4"cos"^(2)(t)/(2)`C. `sqrt(x)=tant,sqrt(y)=sect`D. `x=sqrt(1-sint),y="sin"(t)/(2)+"cos"(t)/(2)` |
|
Answer» Correct Answer - B (2) `x=3cost,y=4sint` Eliminating t, we have `(x^(2))/(9)+(y^(2))/(16)=1` which is an ellipse. `x^(2)-2=2cost andy=4"cos"^(2)(t)/(2)` `or" "y=2(1+cost)` `and" "y=2(1+(x^(2)-2)/(2))` which is a parabola. `sqrt(x)=tant,sqrt(y)=sect` Eliminating t, we have y-x=1 which is a straight line. `x=sqrt(1-sint)` `y="sin"(t)/(2)+"cos"(t)/(2)` Eliminating t, we have `x^(2)+y^(2)=1-sint+1+sint=2` which is a circle. |
|
| 230. |
If the parabola `y=a x^2-6x+b`passes through `(0,2)`and has its tangent at `x=3/2`parallel to the x-axis, then`a=2,b=-2`(b) `a=2,b=2``a=-2,b=2`(d) `a=-2,b=-2`A. a=2, b=-2B. a=2, b=2C. a=-2, b=2D. a=-2, b=-2 |
|
Answer» Correct Answer - B (2) `y=ax^(2)-6x+b` passes through (0,2). Here, `2=a(0^(2))-6(0)+b` `:." "b=2` Also, `(dy)/(dx)=2ax-6` `:." "((dy)/(dx))_(x=3//2)=2a((3)/(2))-6` `or" "3a-6=0` `:." "a=2` |
|
| 231. |
Double ordinate `A B`of the parabola `y^2=4a x`subtends an angle `pi/2`at the focus of the parabola. Then the tangents drawn to the parabolaat `Aa n dB`will intersect at`(-4a ,0)`(b) `(-2a ,0)``(-3a ,0)`(d) none of theseA. (-4a,0)B. (-2a,0)C. (-3a,0)D. none of these |
|
Answer» Correct Answer - A (1) Let `A-=(at^(2),2at),B-=(at^(2),-2at)`. Then `m_(OA)=(2)/(t),m_(OB)=(-2)/(t)` Thus, `((2)/(t))((-2)/(t))=-1` `or" "t^(2)=4` Thus, the tangents will intersect at (-4a,0). |
|
| 232. |
Let `L_(1),L_(2)andL_(3)` be the three normals to the parabola `y^(2)=4ax` from point P inclined at the angle `theta_(1),theta_(2)andtheta_(3)` with x-axis, respectively. Then find the locus of point P given that `theta_(1)+theta_(2)+theta_(3)=alpha` (constant). |
|
Answer» Correct Answer - `y=tanalpha(x-a)` Equation of normal to parabola `y^(2)=4ax` having slope m is `y=mx-2am-am^(3)` This normal passes through point P(h,k) `:." "k=mh-2am-am^(3)` `or" "am^(3)+(2a-h)m+k=0` (1) This equation has three real roots `m_(1),m_(2)andm_(3)`, which are slopes of three normals. Given that `m_(1)=tantheta_(1),m_(2)=tantheta_(2)andm_(3)=tantheta_(3)` From equation (1), we have `m_(1)+m_(2)+m_(3)=0,` `m_(1)m_(2)+m_(2)m_(3)+m_(3)m_(1)=(2a-h)/(a),` `and" "m_(1)m_(2)m_(3)=(-k)/(a)` Now, `theta_(1)+theta_(2)+theta_(3)=alpha` `:." "tan(theta_(1)+theta_(2)+theta_(3))=tanalpha` `rArr" "(m_(1)+m_(2)+m_(3)-m_(1)m_(2)m_(3))/(1-m_(1)m_(2)-m_(2)m_(3)-m_(3)m_(1))=tanalpha` `rArr" "(0+((k)/(a)))/(1-((2a-h)/(a)))=tanalpha` `rArr" "y=tanalpha(x-a)`, which is the required locus. |
|
| 233. |
Find the locus of midpoint of family of chords `lamdax+y=5(lamda` is parameter) of the parabola `x^(2)=20y` |
|
Answer» Equation of family of chord is `(y-5)+lamda(x-0)=0`, which are concurrent at (0,5). The given parabola is `x^(2)-20y=0`. Let M (h,k) be midpoint of chord. Therefore, equation of such chord is `hx-10(y+k)=h^(2)-20k" "(Using T=S_(1))` This chord is passing through the point (0,5). This chord is passing through the point (0,5). `:." "h^(2)=10k-50` So, locus of M (h,k) is `x^(2)=10(y-5)`. |
|
| 234. |
Let `(2,3)` be the focus of a parabola and `x + y = 0` and `x-y= 0` be its two tangents. Then equation of its directrix will beA. `2x - 3y = 0`B. `3x +4y = 0`C. `x +y = 5`D. `12x -5y +1 = 0` |
|
Answer» Correct Answer - A Mirror image of focus in the tangent of parabola lies on its directrix. Here, mirror images of (2,3) in given lines are (3,2) and `(-3,-2)`. `:.` Equation of directrix is `2x - 3y =0`. |
|
| 235. |
The locus of the point of intersection of tangents drawn at the extremities of a normal chord to the parabola `y^2=4ax` is the curveA. x=aB. x=-aC. y=aD. y=-a |
|
Answer» Correct Answer - B From illustration 4, we have `m_(1)m_(2)=a/h` If the tangents are perpndicular, then `m_(1)m_(2)=-1rArra/h=-1rArrh=-a` Hence, the locus of (h, k) is x=-a, which is the directrix of the parabla. |
|
| 236. |
The tangent to `y^(2)=ax` make angles `theta_(1)andtheta_(2)` with the x-axis. If `costheta_(1)costheta_(2)=lamda`, then the locus of their point of intersection isA. `x^(2)=lamda^(2)[(x-a)^(2)+4y^(2)]`B. `x^(2)=lamda^(2)[(+-a)^(2)+y^(2)]`C. `x^(2)=lamda^(2)[(x-a)^(2)+y^(2)]`D. `4x^(2)=lamda^(2)[(x-a)^(2)+y^(2)]` |
|
Answer» Correct Answer - C (3) Let the tangents at `P(at_(1)^(2),2at_(1))andQ(at_(2)^(2),2at_(2))` make angles `theta_(1)andtheta_(2)` with x-axis. Then `"tan"theta_(1)/(t_(1))andtantheta_(2)=(1)/(t_(2))`. Given that `costheta_(1)costheta_(2)=lamda` `rArrsec^(2)theta_(1)sec^(2)theta_(2)=(1)/(lamda^(2))` `rArr(1+tan^(2)theta_(1))(1+tan^(2)theta_(2))=(1)/(lamda^(2))` `rArr(1+(1)/(t_(1)^(2)))(1+(1)/(t_(2)^(2)))=(1)/(lamda^(2))` `rArrlamda^(2)[1+(t_(1)+t_(2))^(2)-2t_(1)t_(2)+t_(1)^(2)t_(2)^(2)]=t_(1)^(2)t_(2)^(2)` Now, point of intersection of tangents is `(at_(1)t_(2),a(t_(1)+t_(2)))-=(h,k)` `rArrlamda^(2)[1+(k^(2))/(a^(2))-(2h)/(a)+(h^(2))/(a^(2))]=(h^(2))/(a^(2))` `rArrx^(2)=lamda^(2)[(x-a)^(2)+y^(2)]` |
|
| 237. |
The angle between tangents to the parabola `y^2=4ax` at the points where it intersects with teine `x-y-a = 0` is `(a> 0)`A. `pi//3`B. `pi//4`C. `pi//6`D. `pi//2` |
|
Answer» Correct Answer - D (4) The coordinates of the focus of the parabola `y^(2)=4ax` are (a,0). the line y-x-a=0 passes through this point. Therefore, it is a focal chord of the parabola. Hence, the tangent intersects at right angle. |
|
| 238. |
PQ is any focal chord of the parabola `y^(2)=8`x. Then the length of PQ can never be less than _________ . |
|
Answer» Correct Answer - 8 (8) The length of focal chord having one extremity `(at^(2),2at)` is `a(t+(1)/(t))^(2)` `because|r+(1)/(t)|ge2` we get `(t+(1)/(t))ge4a=8` or Length of focal chord 8 |
|
| 239. |
From the point (-1,2), tangent lines are to the parabola `y^(2)=4x`. If the area of the triangle formed by the chord of contact and the tangents is A, then the value of `A//sqrt(2)` is ___________ . |
|
Answer» Correct Answer - 8 (8) The chord of contact w.r.t point O(-1,2) is `y=(x-1)" [Using "yy_(1)=2a(x+x_(1))]` Solving y=x-1 with the parabola, we get the point of intersection as `P(3+2sqrt(2),2+2sqrt(2))andQ(3-2sqrt(2),2-2sqrt(2))` `:." "PQ^(2)=32+32=64` `:." "PQ=8` Also, the length of perpendicular from O(-1,2) on PQ is `4//sqrt(2)`. Then the required area of triangle is `A=(1)/(2)xx8xx((4)/(sqrt(2)))=8sqrt(2)` sq. units |
|
| 240. |
If the vertex of a parabola is the point `(-3,0)`and the directrix is the line `x+5=0`, then find its equation.A. `y^(2)=8(x+3)`B. `x^(2)=8(x+3)`C. `y^(2)=-8(x+3)`D. `y^(2)=8(x+5)` |
| Answer» Correct Answer - A | |
| 241. |
The point of intersection of the tangents of the parabola `y^(2)=4x` drawn at the end point of the chord x+y=2 lies onA. x-2y=0B. x+2y=0C. y-x=0D. x+y=0 |
|
Answer» Correct Answer - C (3) Let the point of intersection be `(alpha,beta)`. Therefore, the chord of contact w.r.t. this point is `betay=2x+aalpha` which is the same as x+y=2. Therefore, `alpha=beta=-2` These value satisfy y-x=0. |
|
| 242. |
The length of focal chord to the parabola `y^(2)=12x` drawn from the point (3,6) on is __________ . |
|
Answer» Correct Answer - 12 (12) a=2 Comparing point (3,6) with `3t^(2),6t)`, we have t=1. Then Length of focal `=(t+(1)/(t))^(2)=3(1+1)^(2)=12` |
|
| 243. |
Normals at two points `(x_1y_1)a n d(x_2, y_2)`of the parabola `y^2=4x`meet again on the parabola, where `x_1+x_2=4.`Then `|y_1+y_2|`is equal to`sqrt(2)`(b) `2sqrt(2)`(c) `4sqrt(2)`(d) none of theseA. `sqrt(2)`B. `2sqrt(2)`C. `4sqrt(2)`D. none of these |
|
Answer» Correct Answer - C (3) Normal at point `P(x_(1),y_(1))-=(at_(1)^(2),2at_(1))` meets the parabola at R `(at^(2),2at)`. So, `t=-t_(1)-(2)/(t_(1))` (1) Normal at point `P(x_(2),y_(2))-=(at_(2)^(2),2at_(2))` meets the parabola at R `(at^(2),2at)`. So, `t=-t_(2)-(2)/(t_(2))` (2) From (1) and (2), we get `-t_(1)-(2)/(t_(1))=-t_(2)-(2)/(t_(2))` `:.t_(1)t_(2)=2` Now given that `x_(1)+x_(2)=4`. Therefore, `t_(1)^(2)+t_(2)^(2)=4` `or(t_(1)+t_(2)^(2))=4+4=8` `or|t_(1)+t_(2)|=2sqrt(2)` `or|y_(1)+y_(2)|=4sqrt(2)` |
|
| 244. |
prove that for a suitable point `P`on the axis of the parabola, chord `A B`through the point `P`can be drawn such that `[(1/(A P^2))+(1/(B P^2))]`is same for all positions of the chord. |
|
Answer» Let the point P be (p,0) and the equation of the chord through P be `(x-p)/(costheta)=(y-0)/(sintheta)=r(rinR)` (1) Therefore, `(rcostheta+P,rsintheta)` lies on the parabola `y^(2)=4ax`. So, `r^(2)sin^(2)theta-4arcostheta-4ap=0` (2) If `AP=r_(1)andBP=-r_(2)`, then `r_(1)andr_(2)` are the roots of (2). Therefore, `r_(1)+r_(2)=(4acostheta)/(sin^(2)theta),r_(1)r_(2)=(-4ap)/(sin^(2)theta)` Now, `(1)/(AP^(2))+(1)/(BP^(2))=(1)/(r_(2)^(2))+(1)/(r_(2)^(2))` `=((r_(1)+r_(2))^(2)-2r_(1)r_(2))/(r_(1)^(2)r_(2)^(2))` `=(cos^(2)theta)/(p^(2))+(sin^(2)theta)/(2ap)` Since `(1)/(AP^(2))+(1)/(BP^(2))` should be independent of `theta`, we take p2a. Then, `(1)/(AP^(2))+(1)/(BP^(2))=(1)/(4a^(2))(cos^(2)theta+sin^(2)theta)=(1)/(4a^(2))` Hence, `(1)/(AP^(2))+(1)/(BP^(2))` is independent of `theta` for the positions of the chord if `P-=(2a,0)`. |
|
| 245. |
The endpoints of two normal chords of a parabola are concyclic. Thenthe tangents at the feet of the normals will intersect attangent at vertex of the parabolaaxis of the paraboladirectrix of the parabolanone of theseA. tangent at vertex of the parabolaB. axis of the parabolaC. directrix of the parabolaD. none of these |
|
Answer» Correct Answer - B (2) Let the concyclic points be `t_(1),t_(2),t_(3), and t_(4)` Then, `t_(1)+t_(2)+t_(3)+t_(4)=0` Here, `t_(1)andt_(3)` are the feet of the normal. So, `t_(2)=-t_(1)-(2)/(t_(1))andt_(4)=-t_(3)-(2)/(t_(3))` `:.t_(1)+t_(2)=-(2)/(t_(1))andt_(4)+t_(3)=-(2)/(t_(3))` Therefore, the lies on the intersection of tangents of tangents at `t_(1)andt_(3)` is `(at_(1)t_(3),a(t_(1)+t_(3)))-=(at_(1)t_(3),0)`. This point lies on the axis of the parabola. |
|
| 246. |
The line `x-y=1`intersects the parabola `y^2=4x`at `A`and `B`. Normals at `Aa n dB`intersect at `Cdot`If `D`is the point at which line `C D`is normal to the parabola, then the coordinates of `D`are`(4,-4)`(b) `(4,4)``(-4,-4)`(d) none of theseA. (4,-4)B. (4,4)C. (-4,-4)D. none of these |
|
Answer» Correct Answer - B (2) solving the line y=x-1 and the parabola `y^(2)=4x`, we have `(x-1)^(2)=x` `orx^(2)-6x+1=0` `orx=3pmsqrt(8)` `:.y=2pmsqrt(8)` Suppose point D is `(X_(3),y_(3))`. Then, `y_(1)+y_(2)+y_(3)=0` `or2+sqrt(8)+2-sqrt(8)+y_(3)=0` `ory_(3)=-4` Then `x_(3)=4`. Therefore, the point is (4,4). |
|
| 247. |
Normals are drawn at points `A, B, and C` on the parabola `y^2 = 4x` which intersect at P. The locus of the point P if the slope of the line joining the feet of two of them is 2, is |
|
Answer» The equation of normal at `(r^(2),2t)` is `y=-tx+2t+t^(3)` If it passes through P(h,k), then `t^(3)+t(2-h)-k=0` (1) This equation has roots `t_(1),t_(2)andt_(3)` which are parameters of three feet of normal has three feet of normals on the parabola. From equation (1), we have `t_(1)+t_(2)+t_(3)=0` (2) Given that slope of the line joining the feet of two the normals (say `A(t_(1))andB(t_(2))` is 2. `:." "(2)/(t_(1)+t_(2))=2` `rArr" "t_(1)+t_(2)=1` So, from (2), we have `t_(3)=-1`, which is one of the roots of equation (1). So, from (1) we have -1+h-2-k=0 or x-y-3=0, which is required locus. |
|
| 248. |
Tangents are drawn to the parabola at three distinct points. Prove thatthese tangent lines always make a triangle and that the locus of theorthocentre of the triangle is the directrix of the parabola. |
|
Answer» Let the three point on the parabola be `P(at_(1)^(2),2at_(1)),Q(at_(2)^(2),2at_(2)),R(at_(3)^(2),2at_(3))`. Points of intersection of tangents drawn at these point are `A(at_(2)t_(3),a(t_(2)+t_(3)),B(at_(1)t_(3),a(t_(1)+t_(3)))andC(at_(1)t_(2),a(t_(1)+t_(2)))`. To find the orthocentre of the triangle ABC, we need to find equation of altitudes of the triangle. Slope of BC `=(a(t_(2)-t_(3)))/(at_(1)(t_(2)-t_(3)))=(1)/(t_(1))` Thus, slope of altitude through vertex A is `-t_(1)`. Equation of altitude through A is `y-a(t_(2)+t_(3))=-t_(1)(x-at_(2)t_(3))` `or" "y-a(t_(2)+t_(3))=-t_(1)x+at_(1)t_(2)t_(3)` (1) Similarly, equation of altitude through B is `y-a(t_(1)+t_(3))=-t_(2)xat_(1)t_(2)t_(3)` (2) Subtracting (2) from (1), we get `a(t_(1)-t_(2))=(t_(2)-t_(1))xorx=-a`. Thus , orthocentre lies on the directrix. |
|
| 249. |
The locus of the point of intersection of the perpendicular tangents to the parabola `x^2=4ay` is .A. y=aB. y=-aC. x=aD. x=-a |
| Answer» Correct Answer - B | |
| 250. |
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola :(i) x2 = 16y(ii) x2 = 10y(iii) 3x2 = 8y |
|
Answer» The general form of a parabola: x2 = 4ay ….(1) Focus : F(0,a) Vertex : A(0,0) (at any point A) Equation of the directrix : y + a = 0 Axis: x = 0 Length of latus rectum : 4a (i) x2 = 16y On comparing given equation with (1), we have 4a = 16 => a = 4 Now, Focus : F(0, 4) Vertex : A(0, 0) Equation of the directrix : y + 4 = 0 Axis: x = 0 Length of latus rectum : 4a = 4 x 4 = 16 units (ii) x2 = 10y On comparing given equation with (1), we have 4a = 10 => a = 2.5 Now, Focus : F(0, 2.5) Vertex : A(0, 0) Equation of the directrix : y + 2.5 = 0 Axis: x = 0 Length of latus rectum : 4a = 4 x 2.5 = 10 units (iii) 3x2 = 8y or x2 = 8/3 y On comparing given equation with (1), we have 4a = 8/3 => a = 2/3 Now, Focus : F(0, 2/3) Vertex : A(0, 0) Equation of the directrix : y + 2/3 = 0 or 3y + 2 = 0 Axis: x = 0 Length of latus rectum : 4a = 4 x 2/3 = 8/3 units |
|