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| 251. |
The equation of the directrix of the parabola with vertex at the originand having the axis along the x-axis and a common tangent of slope 2 with thecircle `x^2+y^2=5`is (are)`x=10`(b) `x=20``x=-10`(d) `x=-20`A. x=10B. x=20C. x=-10D. x=-20 |
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Answer» Correct Answer - A::C 1,3 The line y=2x+c ix a tangent to `x^(2)+y^(2)=5`. If `c^(2)=25`, then `c=pm5`. Let the equation of the parabola be `y^(2)=4ax`. Then `(a)/(2)=pm5` `ora=pm10` So, the equation of the parabola is `y^(2)=pm40x`. Also, the equation of the directrices are `x=pm10`. |
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| 252. |
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) `y^(2)=12x` (ii) `y^(2)=10x` (iii)`3y^(2)=8x` |
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Answer» `(i)F(3,0), O(0,0), x+3=0, y=0,12"units"` `(ii)F((5)/(2),0), O(0,0), 2x+5=0, y=0,10"units"` `(iii)F((2)/(3),0), O(0,0), 3x+2=0, y=0,(8)/(3)"units"` |
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| 253. |
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 = -8y(ii) x2 = -18y(iii) 3x2 = -16y |
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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 = -8y On comparing given equation with (1), we have 4a = 8 => a = 2 Now, Focus : F(0, -2) Vertex : A(0, 0) Equation of the directrix : y – 2 = 0 Axis: x = 0 Length of latus rectum : 4a = 4 x 2 = 8 units (ii) x2 = -18y On comparing given equation with (1), we have 4a = 18 => a = 9/2 Now, Focus : F(0, -9/2) Vertex : A(0, 0) Equation of the directrix : y – 9/2 = 0 or 2y – 9 = 0 Axis: x = 0 Length of latus rectum : 4a = 4 x 9/2 = 18 units (iii) 3x2 = -16y Or x2 = -16/3 y On comparing given equation with (1), we have 4a = 16/3 => a = 4/3 Now, Focus : F(0, -4/3) Vertex : A(0, 0) Equation of the directrix : y – 4/3 = 0 or 3y – 4 = 0 Axis: x = 0 Length of latus rectum : 4a = 4 x 4/3 = 16/3 units |
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| 254. |
The locus of the midpoint of the segment joining the focus to a movingpoint on the parabola `y^2=4a x`is another parabola with directrix`y=0`(b) `x=-a``x=0`(d) none of theseA. x = -aB. x = aC. x = 0D. x = a/2 |
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Answer» Correct Answer - C Let `P(at^(2), 2at)` be a moving point on the parabola `y^(2)=4ax` and let S(a, 0) be its focus. Let Q(h, k) be the mid-point of PS. Then, `rArr" "2h=a(k/a)^(2)+a" [On climinating t]"` `rArr" "k^(2)=2a(h-a/2)` Hence, the locus of `"(h, k) is "y^(2)=2a(x-a//2)`. The equation of the directrix of this parabola is `x-a/2=-a/2 rArr x=0` |
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| 255. |
Two straight lines `(y-b)=m_1(x+a)`and `(y-b)=m_2(x+a)`are the tangents of `y^2=4a xdot`Prove `m_1m_2=-1.` |
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Answer» Clearly, both the lines pass through (-a,b), which is a point lying on the directrix of the parabola. Thus, `m_(1)m_(2)=-1`, because tangents drawn from any point on the directrix are always mutually perpendicular. |
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| 256. |
If two tangents drawn from a point P to the parabola y2 = 4x are at rightangles, then the locus of P is(1) `2x""+""1""=""0`(2) `x""=""1`(3) `2x""""1""=""0`(4) `x""=""1`A. 2x-1=0B. x=1C. 2x+1=0D. x=-1 |
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Answer» Correct Answer - D 4 The locus of the perpendicular tangents is the directrix , hance, the locus of P is x=-1. |
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| 257. |
Points A, B, C lie on the parabola `y^2=4ax` The tangents to the parabola at A, B and C, taken in pair, intersect at points P, Q and R. Determine the ratio of the areas of the `triangle ABC` and `triangle PQR` |
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Answer» Correct Answer - 2 2 Let the three points on the parabola `y^(2)=4ax` be `A(t_(1)),B(t_(2)) and C(t_(3))`. Area of triangle ABC, `Delta_(1)=(1)/(2)||{:(1,at_(1)^(2),2at_(1)),(1,at_(2)^(2),2at_(2)),(1,at_(3)^(2),2at_(3)):}||` `=a^(2)||{:(1,t_(1),t_(1)^(2)),(1,t_(2),t_(2)^(2)),(1,t_(3),t_(3)^(2)):}||` `=|a^(2)(t_(1)-t_(2))(t_(2)-t_(3))(t_(3)-t_(1))|` Point of intersection of tangents at A,B and C are `P(at_(1)t_(2),a(t_(1)+t_(2))),` `Q(at_(2)t_(3),a(t_(2)+t_(3))),` `R(at_(3)t_(1),a(t_(3)+t_(1)))`. Area of triangle PQR, `Delta_(2)=(1)/(2)||{:(1,at_(1)t_(2),a(t_(1)+t_(2))),(1,at_(2)t_(3),a(t_(2)+t_(3))),(1,at_(3)t_(1),a(t_(3)+t_(1))):}||` `=(a^(2))/(2)||{:(1,t_(1)t_(2),t_(1)+t_(2)),(1,t_(2)t_(3),t_(2)+t_(3)),(1,t_(3)t_(1),t_(3)+t_(1)):}||` `=(a^(2))/(2)||{:(0,(t_(1)-t_(3))t_(2),t_(1)-t_(3)),(0,(t_(2)-t_(1))t_(3),t_(2)-t_(1)),(1,t_(3)t_(1),t_(3)+t_(1)):}||` (operating `R_(1)toR_(1)-R_(2)andR_(2) to R_(2)-R_(3))` `=(a^(2))/(2)|(t_(1)-t_(3))(t_(2)-t_(1))(t_(2)-t_(3))|` (expanding w.r.t. `C_(1))` (2) From (1) and (2) `(Delta_(1))/(Delta_(2))=2` |
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| 258. |
Find the angle at which the parabolas `y^2=4x`and `x^2=32 y`intersect.A. `tan^(-1)(3//5)`B. `tan^(-2)(4//5)`C. `pi`D. `pi//2` |
| Answer» Correct Answer - A | |
| 259. |
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)x^(2)=16y` (ii)`x^(2)=10y` (iii)`3x^(2)=8y` |
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Answer» `(i)F(0,4), O(0,0), y+4=0, x=0,16"units"` `(ii)F(0,(5)/(2)), O(0,0), 2y+5=0, x=0,10"units"` `(iii)F(0,(2)/(3)), O(0,0), 3y+2=0, x=0,(8)/(3)"units"` |
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| 260. |
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) `y^(2)=-8x` (ii)`y^(2)=-6x` (iii)`5y^(2)=-16x` |
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Answer» `(i)F(-2,0), O(0,0), x=2, y=0,8"units"` `(ii)F((-3)/(2),0), O(0,0), 2x-3=2, y=0,6"units"` `(iii)F((-4)/(5),0), O(0,0), 5x-4=0, y=0,(16)/(5)"units"` |
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| 261. |
If the points (2,3) and (3,2) on a parabola are equidistant from the focus, then the slope of its tangent at vertex isA. 1B. `-1`C. 0D. `oo` |
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Answer» Correct Answer - B Equidistant points from focus are symmetric about axis of parabola. So, tangent at vertex is parallel to the line joining the two points. `:.` Slope `= (3-2)/(2-3) =-1` |
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| 262. |
If the lines `L_1a n dL_2`are tangents to `4x^2-4x-24 y+49=0`and are normals for `x^2+y^2=72 ,`then find the slopes of `L_1`and `L_2dot` |
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Answer» As `L_(1)andL_(2)` are normals to `x^(2)+y^(2)=72`, they must be of the form y=mx. Sloving it with parabola `4x^(2)-4x-24y+49=0`, we have `4x^(2)-4x-24mx+49=0` `or4x^(2)-4x(1+6m)+49=0` Since the lines touch the parabola, D=0 `or16(6m+1)^(2)-16xx49=0` `or6m+1pm7` `i.e.,m=1or-(4)/(3)` |
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| 263. |
Normals are drawn from a point P with slopes `m_1,m_2 and m_3` are drawn from the point p not from the parabola `y^2=4x`. For `m_1m_2=alpha`, if the locus of the point P is a part of the parabola itself, then the value of `alpha` is (a) 1 (b)-2 (c) 2 (d) -1 |
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Answer» Correct Answer - 2 We know equation of normal to `y^(2) = 4ax` is `y = mx - 2am - am^(3)` Thus, equation of normal to `y^(2) = 4x` is `y = mx - 2m - m^(3)` let it passes through (h,k) `implies k = mh - 2m - m^(3)` or `m^(3) + m(2 - h) + k = 0` Here, `m_(1) + m_(2) + m_(3) = 0` `m_(1)m_(2) + m_(2)m_(3) + m_(3)m_(1) = 2 - h` `m_(1)m_(2)m_(3) =- k` where `m_(1)m_(2) = alha` `implies m_(3) = - (k)/(alpha)` it must satisfy Eq. (i) `implies - (k^(3))/(alpha^(3)) - (k)/(alpha) (2 - k) + k = 0` `implies k^(2) = alpha h 2alpha^(2) + alpha^(3)` `implies y^(2) = alpha^(2) x - 2alpha^(2 + alpha^(3)` On comparing with `y^(2 ) = 4x` `implies alpha^(2) = 4` and `-2alpha^(2) + alpha^(3) = 0` `implies alpha = 2` |
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| 264. |
P(-3, 2) is one end of focal chord PQ of the parabola `y^2+4x+4y=0`. Then the slope of the normal at Q isA. `-1//2`B. `2`C. `1//2`D. `-2` |
| Answer» Correct Answer - A | |
| 265. |
If the points A (1,3) and B (5,5) lying on a parabola are equidistant from focus, then the slope of the directrix isA. `(1)/(2)`B. `-(1)/(2)`C. 2D. -2 |
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Answer» Correct Answer - A (1) Since points A and B are equidistant from focus, line AB is perpendicular to axis and so, parallel to directrix. `:.` Slope of directrix = Slope of AB `=(5-3)/(5-1)=(1)/(2)` |
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| 266. |
The circles on the focal radii of a parabola as diameter touch:A) the tangent at the vertexB) the axisC) the directrixD) latus rectumA. the tangent at the vertexB. the axisC. the directrixD. none of these |
| Answer» Correct Answer - A | |
| 267. |
If the normals from any point to the parabola `y^2=4x`cut the line `x=2`at points whose ordinates are in AP, then prove that the slopes of tangents at the co-normalpoints are in GP.A. A.P.B. G.P.C. H.P.D. none of these |
| Answer» Correct Answer - B | |
| 268. |
A parabola is drawn through two given points `A(1,0)` and `B(-1,0)` such that its directrix always touches the circle `x² + y^2 = 4.` Then, The locus of focus of the parabola is=A. `(x^(2))/(4) +(y^(2))/(3) = 1`B. `(x^(2))/(4) +(y^(2))/(5) =1`C. `(x^(2))/(3)+(y^(2))/(4) = 1`D. `(x^(2))/(5)+(y^(2))/(4)=1` |
| Answer» Correct Answer - A | |
| 269. |
Consider a circle with its centre lying on the focus of the parabola, `y^2=2px` such that it touches the directrix of the parabola. Then a point of intersection of the circle & the parabola is:A. `(p//2, +- p)`B. `(p, p//2)`C. `(-p//2, p)`D. `(-p//2, -p)` |
| Answer» Correct Answer - A | |
| 270. |
Tangent `P Aa n dP B`are drawn from the point `P`on the directrix of the parabola `(x-2)^2+(y-3)^2=((5x-12 y+3)^2)/(160)`. Find the least radius of the circumcircle of triangle `P A Bdot` |
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Answer» We have equation of parabola, `sqrt((x-2)^(2)+(y-3)^(2))=(|5x-12y+3|)/(sqrt(5^(2)+(-12)^(2)))` Focus of the parabola is (2,3) and directrix is 5x-12y+3=0. Now, tangents drawn to parabola from point P on the directrix are perpendicular and the corresponding chord of contact AB focal chord which is diameter of the circumcircle of the triangle PAB. So, least value of diameter is latus rectum. Here, `L.R=2xx` Distance of focus from directrix `=2xx(|10-36+3|)/(13)=(23)/(13)` So, required radius `=(46)/(13)` |
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| 271. |
The length of the latus rectum of the parabola whose focus is `((u^2)/(2g)sin2alpha,-(u^2)/(2g)cos2alpha)`and directrix is `y=(u^2)/(2g)`is`(u^2)/gcos^2alpha`(b) `(u^2)/gcos^2 2alpha``(2u^2)/gcos^2 2alpha`(d) `(2u^2)/gcos^2alpha`A. `u^(2)/gcos^(2)alpha`B. `u^(2)/gcos2alpha`C. `(2u^(2))/gcos2alpha`D. `(2u^(2))/gcos^(2)alpha` |
| Answer» Correct Answer - D | |
| 272. |
Find the vertex, focus and directrix of the parabola `x^(2)=2(2x+y)`. |
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Answer» Correct Answer - `"Vertex"-=(2,-2),"Focus"-=(2,-3//2),"Directrix":y=-(5)/(2)` We have `x^(2)=2(2x+y)` `:." "x^(2)-4x+4=2(y+2)` `rArr" "(x-2)^(2)=2(y+2)` Comparing with `(x-h)^(2)=2(y-k)`, we get Vertex of the parabola is (2,-2). Length of latus rectum =4a=2 Axis of the parabola is x=2. Also, parabola is concave upward. Focus lies at distance a units above the vertex on the axis. So, focus is `(2,-2+(1)/(2))-=(2,-(3)/(2))`. Directrix is at distance a units below the vertex and perpendicular to the axis. Therefore, directrix is `y=-2-(1)/(2)ory=-(5)/(2)`. |
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| 273. |
If P be a point on the parabola `y^2=3(2x-3)` and M is the foot of perpendiculardrawn from the point P on the directrix of the parabola, then length of each sides of an equilateral triangle SMP(where S is the focus of the parabola), isA. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - C (3) `y^(2)=6(x-(3)/(2))` The equation of directrix is `x-(3)/(2)=-(3)/(2)orx=0` Vertex is `((3)/(2),0)` and focus of P be `((3)/(2)+(3)/(2)t^(2),3t)`. Then the coordinates of M are (0,3t). Therefore, `MS=sqrt(9+9t^(2))` `MP=(3)/(2)+(3)/(2)t^(2)` `or9+9t^(2)=((3)/(2)+(3)/(2)t^(2))^(2)=(9)/(4)(1+t^(2))^(2)` `or4=1+t^(2)` `:.` Length of side = 6 |
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| 274. |
the tangent drawn at any point `P` to the parabola `y^2= 4ax` meets the directrix at the point `K.` Then the angle which `KP` subtends at the focus isA. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - D | |
| 275. |
Show that `xcosalpha+asin^2alpha=p`touches the parabola `y^2=4a x`if `pcosalpha+asin^2alpha=0`and that the point of contact is `(atan^2alpha,-2atanalpha)dot` |
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Answer» The given line is `xcosalpha+ysinalpha=p` `ory=-xcotalpha+p" cosec "alpha` `:.m=-cotalphaandc=p" cosec "alpha` Since the given line touches the parabola, we have `c=(a)/(m)` or cm=a or `(pcosalpha+asin^(2)alpha=0)` `orpcosalpha+asin^(2)alpha=0` The point of contact is `((a)/(cot^(2)alpha),(2a)/(cotalpha))-=(atan^(2)alpha,-2atanalpha)` |
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| 276. |
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.A. an ellipe which is not a circleB. a hyperbolaC. a parabolaD. a circle |
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Answer» Correct Answer - C Let C(h, k) be the centre of the circle touching x=axis and the circle `x^(2)+y^(2)-8x-8y-4=0` externally. The centre `C_(1)` of this circle has coordinates (4, 4) and its radius is 6. The circles with centres at C and `C_(1)` touch each other externally. `:." "C C_(1)=|k|+6` `rArr" "sqrt((h-4)^(2)+(k-4)^(4))=|k|+6` `rArr" "(h-k)^(2)+(k-4)^(2)=(|k|+6)^(2)` `rArr" "(h-4)^(2)-8k=12|k|+20` `rArr" "(h-k)^(2)=20k+20, if kgt0` `"or, "(h-k)^(2)=-4k+20, ifklt0` Hence, the locus of (h, k) is `(x-4)^(2)=20(y+1)or,(x-4)^(2)=-4(y-5)` which is a parabola. |
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| 277. |
If the focus of a parabola is (2, 3) and its latus rectum is 8, then find the locus of the vertex of the parabola. |
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Answer» Latus rectum =4a=8. `:." "a=2`. Now, the vertex lies at distance a units from the focus. Let the vertex be (x,y). `:." "(x-2)^(2)+(y-3)^(2)=4` which is the equation of locus of the vertex. |
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| 278. |
If the length of the latus rectum of x2 = 4ky is 8, find the value of k. |
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Answer» Compare x2 = 4ky with x2 = 4ay. We get 4k = 4a ⇒ a = k Given length of LR = 8 4a = 8 4k = 8 k = 2. |
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| 279. |
The vertex of a parabola is (2, 2) and the coordinats of its twoextremities of latus rectum are `(-2,0)`and (6, 0). Then find the equation of the parabola. |
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Answer» Correct Answer - `(x-2)^(2)=-8(y-2)` Focus is midpoint of the extremities of latus rectum. Thus, focus is (2,0). Distance between focus and vertex is a=2. Also, axis of the parabola is x=2 and parabola is concave downward as focus lies below the vertex. Therefore, using equation `(x-h)^(2)=-4a(y-k)`, required equation of parabola is `(x-2)^(2)=-8(y-2)` |
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| 280. |
A circle is drawn to pass through the extremities of the latus rectumof the parabola `y^2=8xdot`It is given that this circle also touches the directrix of theparabola. Find the radius of this circle. |
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Answer» Correct Answer - 4 The extremities of the latus rectum are (2,4) and (2,-4). Since any circle drawn with any focal chord as its diameter touches the directrix, the radius of the circle is 2a=4 (sincea=2). |
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| 281. |
The chords of contact of the pairs of tangents drawn from each point on the line `2x +y=4` to the parabola `y^2=-4x` pass through the pointA. `(2, -1)`B. `(1//2, 1//4)`C. `(-1//2,-1//4)`D. `(-2, 1)` |
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Answer» Correct Answer - D Let P(t, -2t) be any point on 2x+y=4. The chord of contact of tangents drawn from p to `y^(2)=-4x` is `(4-2t)y=-2(x+1)` `or, (2x+4y)-2t(y-1)=0` Clearly, it represents a line passing through the point of intersection of the lines 2x+4y=0 and y-1=0 i.e.(-2, 1). |
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| 282. |
If `aa n dc`are the lengths of segments of any focal chord of the parabola `y^2=b x ,(b >0),`then the roots of the equation `a x^2+b x+c=0`arereal and distinct(b) real and equalimaginary(d) none of theseA. real and distinctB. real and equalC. imaginaryD. none of these |
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Answer» Correct Answer - C (3) The latus rectum of `y^(2)=2bx` is 2b. The semi-latus rectum is the HM of the segments of focal chord. Then, `2x(t_(1)t_(2)+1)+2a(t_(1)t_(2)+1)+0` `or(x+a)(1+t_(1)t_(2))=0` `orx+a=0` Now, for `ax^(2)+bx+c=0` `D=b^(2)-4ac=((2ac)/(a+c))^(2)-4ac` `=-4ac((a^(2)+c^(2)-ac)/((a+c)^(2)))lt0` Hence, the roots are imaginary. |
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| 283. |
AB, AC are tangents to a parabola `y^(2)=4ax`. If `l_(1),l_(2),l_(3)` are the lengths of perpendiculars from A, B, C on any tangent to the parabola, thenA. `l_(1), l_(2), l_(3)` are in GPB. `l_(2), l_(1), l_(3)` are in GPC. `l_(3), l_(1), l_(2)` are in GPD. `l_(3), l_(2), l_(1)` are in GP |
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Answer» Correct Answer - B::C Let the coordinates of B C be `(at_(1)^(2), 2at_(1))` and `(at_(1)t_(2), a(t_(1)+t_(2))` respectively. Then, the coordinates of A are `(at_(1)t_(2), a(t_(1)+t_(2))` The equation of any tangent to `y^(2)=4ax` is `ty=x+aty^(2)` `:." "l_(1)=(at_(1)t_(2)-a(t_(1)-t_(2))t+at^(2))/(sqrt(1+t^(2)))` Clearly, `l_(2)l_(3)=l_(1)^(2)`. Therefore, `l_(2)l_(1)l_(3)` are om G.P. |
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| 284. |
Find the equation of line which is normal to the parabola `x^(2)=4y` and touches the parabola `y^(2)=12x`. |
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Answer» Normal to parabola `x^(2)=4y` having slope m is `y=mx+2+(1)/(m^(2))` (1) It is tangent to `y^(2)=12x`. Now, tangent to above parabola having slope m is `y=mx+(3)/(m)` (2) Comparing (1) and (2), we get `(1)/(m^(2))+2=(3)/(m)` `rArr" "2m^(2)-3m+1=0` `rArr" "(2m-1)(m-1)=0` `rArr" "m=(1)/(2)orm=1` Therefore, equations of lines are 2y=x+12 or y=x+3. |
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| 285. |
Given a parabola `y^(2) = 4ax` and the points `A(a t^(2),2at), B(a t^(-2),2a t^(-1)), C ((4a)/(t^(2)),(4a)/(t))D (a(t+(2)/(t))^(2),-2a (t+(2)/(t)))` choose all the correct alternative.A. AB is a focal chordB. AD is a normal chordC. Normals at A,C intersect on the parabolaD. Tangents at A,B intersect at `90^(@)` on the directrix. |
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Answer» Correct Answer - A::B::C::D AB is focal chord, tangents at which to the parabola intersect on dirextrix. AD is normal chord of the parabola. |
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