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| 101. |
Find the vertex, focus, axis, directrix and lotus – rectum of the following parabolas(i) y2 = 8x(ii) 4x2 + y = 0(iii) y2 – 4y – 3x + 1 = 0(iv) y2 – 4y + 4x = 0(v) y2 + 4x + 4y – 3 = 0 |
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Answer» (i) y2 = 8x Given: Parabola = y2 = 8x Now by comparing with the actual parabola y2 = 4ax Then, 4a = 8 a = 8/4 = 2 So, the vertex is (0, 0) The focus is (a, 0) = (2, 0) The equation of the axis is y = 0. The equation of the directrix is x = – a i.e., x = – 2 The length of the latus rectum is 4a = 8. (ii) 4x2 + y = 0 Given: Parabola => 4x2 + y = 0 Now by comparing with the actual parabola y2 = 4ax Then, 4a = 1/4 a = 1/(4 × 4) = 1/16 So, the vertex is (0, 0) The focus is = (0, -1/16) The equation of the axis is x = 0. The equation of the directrix is y = 1/16 The length of the latus rectum is 4a = 1/4 (iii) y2 – 4y – 3x + 1 = 0 Given: Parabola y2 – 4y – 3x + 1 = 0 y2 – 4y = 3x – 1 y2 – 4y + 4 = 3x + 3 (y – 2)2 = 3(x + 1) Now by comparing with the actual parabola y2 = 4ax Then, 4b = 3 b = 3/4 So, the vertex is (-1, 2) The focus is = (3/4 – 1, 2) = (-1/4, 2) The equation of the axis is y – 2 = 0. The equation of the directrix is (x – c) = -b (x – (-1)) = -3/4 x = -1 – 3/4 = -7/4 The length of the latus rectum is 4b = 3 (iv) y2 – 4y + 4x = 0 Given: Parabola y2 – 4y + 4x = 0 y2 – 4y = – 4x y2 – 4y + 4 = – 4x + 4 (y – 2)2 = – 4(x – 1) Now by comparing with the actual parabola y2 = 4ax => (y – a)2 = – 4b(x – c) Then, 4b = 4 b = 1 So, the vertex is (c, a) = (1, 2) The focus is (b + c, a) = (1-1, 2) = (0, 2) The equation of the axis is y – a = 0 i.e., y – 2 = 0 The equation of the directrix is x – c = b x – 1 = 1 x = 1 + 1 = 2 Length of latus rectum is 4b = 4 (v) y2 + 4x + 4y – 3 = 0 Given: The parabola y2 + 4x + 4y – 3 = 0 y2 + 4y = – 4x + 3 y2 + 4y + 4 = – 4x + 7 (y + 2)2 = – 4(x – 7/4) Now by comparing with the actual parabola y2 = 4ax => (y – a)2 = – 4b(x – c) Then, 4b = 4 b = 4/4 = 1 So, The vertex is (c, a) = (7/4, -2) The focus is (- b + c, a) = (-1 + 7/4, -2) = (3/4, -2) The equation of the axis is y – a = 0 i.e., y + 2 = 0 The equation of the directrix is x – c = b x – 7/4 = 1 x = 1 + 7/4 = 11/4 Length of latus rectum is 4b = 4. |
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| 102. |
For the parabola, y2 = 4px find the extremities of a double ordinate of length 8p. Prove that the lines from the vertex to its extremities are at right angles. |
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Answer» Given: The parabola, y2 = 4px and a double ordinate of length 8p. Let AB be the double ordinate of length 8p for the parabola y2 = 4px. Now, let us compare to the actual parabola, y2 = 4ax Then, axis is y = 0 vertex is O(0, 0). We know that double ordinate is perpendicular to the axis. So, let us assume that the point at which the double ordinate meets the axis is (x1, 0). Then the equation of the double ordinate is y = x1. It meets the parabola at the points (x1, 4p) and (x1, – 4p) as its length is 8p. Now, let us find the value of x1 by substituting in the parabola. (4p)2 = 4p(x1) x1 = 4p. The extremities of the double ordinate are A(4p, 4p) and B(4p, – 4p). Assume the slopes of OA and OB be m1 and m2. Let us find their values. m1 = (4p – 0)/(4p – 0) = 4p/4p = 1 m2 = (4p – 0)/(-4p – 0) = 4p/-4p = -1 = – 1 The product of slopes is -1. So, the lines OA and OB are perpendicular to each other. Hence the extremities of double ordinate make right angle with the vertex. |
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| 103. |
The point on the parabola `y^(2) = 8x` at which the normal is inclined at `60^(@)` to the x-axis has the co-ordinates asA. `(6,-4sqrt(3))`B. `(6,4sqrt(3))`C. `(-6,-4sqrt(3))`D. `(-6,4sqrt(3))` |
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Answer» Correct Answer - A Slope of normal `= tan 60^(@) = sqrt(3)` `:.` Point of normal is `(2(sqrt(3))^(2), -2(2)sqrt(3))` or `(6,-4sqrt(3))` |
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| 104. |
The area between the parabola `y^2 = 4x` , normal at one end of latusrectum and X-axis in sq.units isA. `60^(@)`B. less then `60^(@)`C. more then`60^(@)`D. less then `45^(@)` |
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Answer» Correct Answer - C Let `y+t_(1)x=2t_(1)+t_(1)^(3)` be a normal at `P(t_(1)^(2), 2t_(1))` to the parabola `y^(2)=4x`. Suppose it cuts the parabola again at `Q(t_(2)^(2), 2t_(2))`. Then, `t_(2)=-t_(1)-2/t_(1)` The equation of the normal at Q is `y+t_(2)x=2t_(2)+t_(2)^(3)` Suppose it makes an angle `theta` with the positive direction of x-axis. `:." "tantheta=-t_(2)` `rArr" "tantheta=t_(1)+2/t_(1)` `rArr" "|tantheta|=|t_(1)+2/t_(1)|ge2sqrt(t_(1)xx2/t_(1))" "["Using:"AMgeGM]` `rArr" "|tantheta|ge2sqrt2rArr0gt60^(@)` |
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| 105. |
If the equation `lambdax^2+ 4xy + y^2 + lambdax + 3y + 2 = 0` represent a parabola then find `lambda`.A. -4B. 4C. 0D. none of these |
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Answer» Correct Answer - C Comparing the given equation with the general equation of second degree i.e. `ax^(2)+2hxy+by^(2)+2gx+2fy+c=0`, we have ltbtgt `a-lamda, b=1, c~~2, f=3/2, g=lamda/2 " and "h-2` `:." "Delta=abc+2fgh-of^(2)-bg^(2)-ch^(2)` `rArr" "Delta=2lamda=3lamda-(9lamda)/4-(lamda^(2))/4-8=-1/4(lamda^(2)-11lamda+32)` and, `h^(2)-ab=4-lamda`. If the given equation represents a parabola, we must have `Deltaneh^(2)" and "h^(2)=ab` Clearly, `lamda=4` sarisfies these two conditions. |
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| 106. |
The equation `x^(2)+4xy+4y^(2)-3x-6y-4=0` represents aA. circleB. parabolaC. a pair of straght linesD. none of these |
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Answer» Correct Answer - C Here, a=1, b=4, c=-4, f=-3, g=-3/2 and h=2. `:." "Delta=abc+2fgh-af^(2)-bg^(2)-ch^(2)=-16+18-9-9+16=0" and, " h^(2)-ab=4-4=0` So, the given equation represents a pair of parallel straight lines. |
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| 107. |
If three parabols touch all the lines `x = 0, y = 0` and `x +y =2`, then maximum area of the triangle formed by joining their foci isA. `sqrt(3)`B. `sqrt(6)`C. `(3sqrt(3))/(4)`D. `(3sqrt(3))/(2)` |
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Answer» Correct Answer - D Let ABC be a triangle whose sides are `x =0, y =0` and `x +y = 2`. Since, the parabols touch the side, so foci must lie on circumcircle of the triangle ABC whose radius is `sqrt(2)`. Now, foci form a triangle of maximum area. Hence, foci must be the vertices of an equilateral triangle inscribed in the circumcircle. So, area `= (sqrt(3))/(4) (sqrt(3)sqrt(2))^(2) = (3sqrt(3))/(2)`. |
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| 108. |
If the parabols `y^(2) = 4kx (k gt 0)` and `y^(2) = 4 (x-1)` do not have a common normal other than the axis of parabola, then `k in`A. `(0,1)`B. `(2,oo)`C. `(3,oo)`D. `(0,oo)` |
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Answer» Correct Answer - A::B::C If the parabolas have a common normal of slope `m(m ne 0)` then it is given by `y = mx - 2km -km^(3)` and `y = m(x-1) -2m -m^(3) = mx -3m -m^(3)` `rArr 2km + km^(3) = 3m + m^(3)` `rArr m = 0, m^(2) =(3-2k)/(k-1)`. If `m^(2) lt 0` then the only common normal is the axis. `rArr (3-2k)/(k-1) lt 0` `rArr (k-1) (2k-3) gt 0` `k gt (3)/(2)` or `k lt 1` and `k gt 0` |
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| 109. |
Let `y^(2) -5y +3x +k = 0` be a parabola, thenA. its latus rectum is least when `k = 1`B. its latus rectum is independent of kC. the line `y = 2x +1` will touch the parabola if `k = (73)/(16)`D. `y = (5)/(2)` is the only normal to the parabola whose slope is zero |
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Answer» Correct Answer - B::C::D The equation to the parabola can be written as: `(y-(5)/(2))^(2) =- 3 (x-(25-4k)/(12))` The length of the latus rectum is `3.y = 2x+1` is a tangent. So, the quadratic equation `(2x+1)^(2) - 5 (2x+1) +3x +k =0` or `4x^(2) -3x +k-4 =0` must have equal roots. Now roots are equal if `b^(2) = 4ac` `rArr 9 = 16(k-4) rArr k =(73)/(16)` `y = (5)/(2)` is normal at the vertex which has slope 0. |
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| 110. |
Does equation `(5x-5)^(2)+(5y+10)^(2)=(3x+4y+5)^(2)` represents a parabola ? |
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Answer» We have `(5x-5)^(2)+(5y+10)^(2)=(3x+4y+5)^(2)` `:." "sqrt((x-1)^(2)+(y+2)^(2))=(|3x+4y+5|)/(5)` L.H.S = distance of variable point P(x,y) from fixed point (1,-2) R.H.S. = distance of point P(x,y) from the fixed line `3x+4y+5=0` But point (1,-2) satisfies the line `3x+4y+5=0`. So, equation does not represent a parabola. |
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| 111. |
Find the value of `P`such that the vertex of `y=x^2+2p x+13`is 4 units above the x-axis. |
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Answer» The given parabola is `y=x^(2)+2px+13` `ory=(x+p)^(2)+13-p^(2)` `ory-(13-p^(2))=(x+p)^(2)` So, the vertex is `(-p,13-p^(2))`. Since the parabola is at a distance of 4 units above the x-axis, `13-p^(2)=4` `orp^(2)=9` `orp=pm3` |
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| 112. |
Find the equation of the parabola which has axis parallel to the y-axisand which passes through the points `(0,2),(-1,0),a n d(1,6)dot` |
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Answer» General equation of parabola having axis parallel to y-axis is `y=Ax^(2)+Bx+C` It passes through the points (0,2), (-1,0) and (1,6). Then we have, C=2 (1) A-B+C=0 (2) A+B+C=6 (3) Solving (1), (2) and (3), we get C=2, A=1 and B=3. Therefore, equation of parabola is `y=x^(2)+3x+2`. |
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| 113. |
let `P` be the point `(1, 0)` and `Q` be a point on the locus `y^2= 8x`. The locus of the midpoint of `PQ` isA. `x^(2)-4y=2=0`B. `x^(2)+4y+2-0`C. `y^(2)+4x+2=0`D. `y^(2)-4x+2=0` |
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Answer» Correct Answer - D Let R(h, k) be the mid-point of PQ and the coordintes at Q be `(alpha, beta)`. Then, `(alpha+1)/2=h" and "(beta+0)/2=krArralpha=h-1" and "beta=2k` Since `Q (alpha, beta)` lies on the parabola `y^(2)=8x.` `:." "beta^(2) -8alpharArr4k^(2)=8(2h-1)rArrk^(2)=4k-2` Hence, the locus of (h, k) is `y^(2)=4x-2` |
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| 114. |
If `x=m x+c`touches the parabola `y^2=4a(x+a),`then`c=a/m`(b) `c=a m+a/m``c=a+a/m`(d) none of theseA. `c=(1)/(m)`B. `c=am+(a)/(m)`C. `c=a+(a)/(m)`D. none of these |
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Answer» Correct Answer - B (2) The equation of tangent to the given parabola having slope m is `y=m(x+a)+(a)/(m)` `ory=mx+am+(a)/(m)` Comparing (1) with y=mx+c, we have `c=am+(a)/(m)` |
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| 115. |
If the line y=mx+c touches the parabola `y^(2)=4a(x+a)`, thenA. `c=a+a/m`B. `c=am+a/m`C. `c=am+a`D. none of these |
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Answer» Correct Answer - B The equation of the parabola is `y^(2)=4a(x+6)" …(i)"` The equation of the line `y = mx + c` can be written as `y=m(x+a)+c-am" …(ii)"` we know that the line` y=mc+lamda` touches the parabola `y^(2)=4ax`, if `lamda=a/m`. Therefore, line (ii) will touch the parabola (i), if `c-am=a/mrArrc=am+a/m` ALITER The y=mx+c will touch the parabola `y^(2)=4a(x+a)` if the quadratic equation `(mx+c)^(2)=4a(x+a)` has equal roots i.r. its discireminant is equal to zero. `:." "4(mc-2a)^(2)-4m^(2)(c^(2)-4a^(2))rArrc=am+a/m` |
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| 116. |
Find the point on the curve `y^2=a x`the tangent at which makes an angle of 45^0 with the x-axis.A. (a/2, a/4)B. (-a/2, a/4)C. (a/4, a/2)D. (-a/4, a/2) |
| Answer» Correct Answer - C | |
| 117. |
Find the equations of the tangent and the normal to the given curve at the indicated point : `y^(2) = 4ax " at " ((a)/(m^(2)), (2a)/(m))`A. `y=mx-2am-am^(3)`B. `m^(3)y=m^(2)x-2am^(2)-a`C. `m^(3)y=2am^(2)-m^(2)x+a`D. none of these |
| Answer» Correct Answer - C | |
| 118. |
If `y=2x+k` is a tangent to the curve `x^(2)=4y`, then k is equal toA. 4B. 43467C. -4D. `-1//2` |
| Answer» Correct Answer - C | |
| 119. |
The mid-point of the line joining the common points of the line `2x-3y+8=0" and "y^(2)=8x,` isA. (3, 2)B. (5, 6)C. (4, -1)D. (2, -3) |
| Answer» Correct Answer - B | |
| 120. |
The point P on the parabola `y^(2)=4ax` for which | PR-PQ | is maximum, where R(-a, 0) and Q (0, a) are two points,A. (a, 2a) or (a, -2a)B. (a, -2a)C. (4a, 4a)D. (4a, -4a) |
| Answer» Correct Answer - A | |
| 121. |
Tangents PQ and PR are draqn to the parabola `y^(2)=20(x+5)" and "y^(2)=60(x+15)` respectively such that `/_RPQ=pi/2,` the locus of point P, isA. x+10=0B. x+30=0C. x+40=0D. none of these |
| Answer» Correct Answer - D | |
| 122. |
If `m` be the slope of common tangent of `y = x^2 - x + 1` and `y = x^2 – 3x + 1`. Then `m` is equal toA. 16B. 7C. 9D. none of these |
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Answer» Correct Answer - B Let y = mx + c be the common tangent to the two parabolas. Then, the equations. `xm+c=x^(2)-3x+10" and "mx+c=x^(2)-x+1` must have equal roots i.e. `x^(2)-x(3+m)+10-c=0" and "x^(2)-x(1+m)=4(1-c)` must have equal roots `:." "(3+m)^(2)=4(10-x)" and "x^(2)-x(1+m)+1-c=0` `rArr" "(3+m)^(2)-(1+m)^(2)=36rArrm=7`. |
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| 123. |
The slope of the line touching both the parabolas `y^2=4x and x^2=−32y` isA. `1//2`B. `3//2`C. `1//8`D. `2//3` |
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Answer» Correct Answer - A 1 Equation of tangent to `y^(2)=4x` at A `(t^(2),2t)` is `yt=x+t^(2)` This is tangent to `x^(2)+32=0` `rArrx^(2)+32((x)/(t)+t)=0rArrx^(2)+(32)/(t)x+32t=0` Above equation must have eqaution roots `rArr((32)/(t))^(2)-4(32t)=0` `rArr32((32)/(t^(2))-4t)=0` `rArrt^(3)=8rArrt=2` `rArr" Slope of tangent is "(1)/(t)=(1)/(2)`. |
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| 124. |
The slope of the line touching both the parabolas `y^2=4x and x^2=−32y` is (a) `1/2` (b) `3/2` (c) `1/8` (d) `2/3`A. 43473B. 43499C. 43467D. 43526 |
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Answer» Correct Answer - C The equation of any tangent to `y^(2)=4x` is `y=mx+1/m"or, "x=1/my-1/m^(2)`. If it touches `x^(2)=-32y,` then As the equation of any tangent to `x^(2)=-32y`, is of the form `x=my-8/m.` `:." "-1/m^(2)=-8/("1/m")rArrm=1/2` ALITER The equation of any tangent to `y^(2)=4x" is "y =mx+1/m`. If it touches `x^(2)=-32y," then "x^(2)=-32(mx+1/2)` must have equal roots. `:." "(32m)^(2)=4((32)/(m))rArr=m=1/2` |
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| 125. |
The equation of the line touching both the parabolas `y^(2)=4xandx^(2)=-32y` is ax+by+c=0. Then the value of a+b+c is ___________ . |
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Answer» Correct Answer - 3 (3) The equation of tangent to parabola `y^(2)=4x` is `y=mx+(1)/(m)` Since (1) is also the tangent of `x^(2)=-32y`, we have `x^(2)=-32(mx+(1)/(m))` `orx^(2)+32mx+(32)/(m)=0` The above equation must have equal roots. Hence, its discriminant must be zero. Therefore, `(32m)^(2)=4xx1xx(32)/(m)` `i.e.," "m^(3)=(1)/(8)or=(1)/(2)` From (1), `y=(x)/(2)+2` `orx-2y+4=0` |
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| 126. |
An equilateral triangle is inscribed in theparabola `y^2=4a x`whose vertex is at of the parabola. Find the length ofits side.A. `4asqrt3`B. `2asqrt3`C. `16asqrt3`D. `8asqrt3` |
| Answer» Correct Answer - D | |
| 127. |
If the length of the latus rectum of y2 = 8kx is 4 find k |
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Answer» Compare y2 = 8kx with y2 = 4ax We get 4a = 8k ⇒ a = 2k Given length of LR = 4 4a = 4 ⇒ 8k = 4 ⇒ k = \(\frac{1}{2}\) |
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| 128. |
Circle described on the focal chord as diameter touches the tangent at the vertexA. the axisB. the tangent at the vertexC. the directrixD. none of these |
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Answer» Correct Answer - B Let S(a, 0) be the focus of the parabola `y^(2)=4ax`, and `P(at^(2), 2at)` be a point on it. Then equation of a circle on SP as diameter is `(x-a)(x-at^(2))+(y-0)(y-2at)=0` It meets y-axis at x = 0 `:." "y^(2)-2"at y"+a^(2)t^(2)=0rArr(y-t)^(2)=0` This shows that y-axis meets the circle in two coincident points. Hence, the circle touches the tangent at the vertex. |
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| 129. |
If the line `y=3x+c`touches the parabola `y^2=12 x`at point `P`, then find the equation of the tangent at point `Q`where `P Q`is a focal chord. |
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Answer» Correct Answer - `x+3y+27=0` Line `(1)/(3)y=x+(c)/(3)` touches the parabola `y^(2)=12x`. Let us compare this line `ty=x+at^(2)orty=x+3t^(2)`. So, we have `t=(1)/(3)` which is parameter of point P. Since PQ is focal chord, parameter of point Q is -3. Therefore, equation of tangent at Q is `(-3)y=x+3(-3)^(2)` `or" "x+3y+27=0` |
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| 130. |
If the line `y=m x+1`is tangent to the parabola `y^2=4x ,`then find the value of `mdot` |
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Answer» We know, equation of a tangent for a parabola, `y^2 = 4ax` is, `y = mx+a/m` Comparing it with given equation, `y = mx+1` `a/m = 1` `=> m = a->(1)` In the give parabola, `y^2 = 4x`, `a = 1` `:. m = a = 1` So, the value of `m` is `1`. |
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| 131. |
If lx + my + n = 0 is tangent to the parabola `x^(2)=y`, themA. `t^(2)=2mn`B. `i=4m^(2)n^(2)`C. `m^(2)=4/n`D. `l^(2)=4mn` |
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Answer» Correct Answer - D We know that x = my+x touche the parabola `x^(2)=4ay`, if `c=a/m`, Therefore , lx+my+n=0 or `x=-m/ly+-n/l` touches `x^(2)=y`, if `(-n)/l=((1//4))/((-m//l))rArr(mn)/l^(2)=1/4rArrl^(2)=4mn` `Delta"LITER"` If lx + my + n = 0 touches the parabola `x^(2)=y`, then `lx+mx^(2)+n=0` must have equal roots. `:." "l^(2)=4mn" [On equating discriminant to zero]"` |
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| 132. |
If the line `y=m x+1`is tangent to the parabola `y^2=4x ,`then find the value of `mdot`A. 1B. 2C. 4D. 3 |
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Answer» Correct Answer - A If y = mx + 1 touches the parabola `y^(2)=4x`, then the equation must have equal roots. `:." "4(m-2)^(2)=4m^(2)rArrm=1` ALITER We know tha t = mx + c touches `y^(2)=4ax,` if c = a/m. `:." y = mx + 1 will touch "y^(2)=4x,` if `1=1/mrArrm=1" "["Using : c"=a/m]` |
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| 133. |
If the line x + y - 1 = 0 touches the parabola `y^(2)=kx,` thn the value of k, isA. 4B. -4C. 2D. -2 |
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Answer» Correct Answer - B We have, `x+y-1=0rArry=-x+1` This will touch the parabola `y^(2)=kx,` if `1=(k//4)/-1" [putting c=1, a=k/4 and m = -1 in c=a/m]"` `rArr" "k=-4` |
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| 134. |
The harmonic mean of the segments of a focal chord of the parabola `y^(2)=16ax,` isA. 2aB. 4aC. 8aD. 16a |
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Answer» Correct Answer - C The harmonic mean of the segments of a focal chord of the parabola is its semi-latusrectum i.e. 8a. |
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| 135. |
The ratio in which the line segment joining the point (4, -6) and (3, 1) si divided by the parabola `y^(2)=4ax` isA. `(-20+-sqrt(155))/(11):1`B. `(-2+-sqrt(155))/(11):1`C. `-20+-2sqrt(155):11`D. `-20+-sqrt(155):11` |
| Answer» Correct Answer - C | |
| 136. |
Equation of the chord of the parabola `y^2 = 8x` which is bisected at the point `(2,-3)` isA. `4x+3y+1=0`B. `2x+3y+5=0`C. `3x+4y+6=0`D. `2x-3y-12=0` |
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Answer» Correct Answer - A The equation of the required chord id given by `S=TrArr9-8xx2=-3y-4(x+2)rArr4x+3y+1=0` |
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| 137. |
Let a, r, s, t be non-zero real numbers. Let `P(at^(2),2at),Q(ar^(2),2ar)andS(as^(2),2as)` be distinct points on the parabola `y^(2)=4ax`. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K the point (2a,0). If st=1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate isA. `((t^(2)+1)^(2))/(2t^(3))`B. `(a(t^(2)+1)^(2))/(2t^(3))`C. `(a(t^(2)+1)^(2))/(t^(3))`D. `(a(t^(2)+2)^(2))/(t^(3))` |
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Answer» Correct Answer - B 2 Tangent at `P:ty=x+at^(2)ory=(x)/(t)+at` Normal at `S:y+(x)/(t)=(2a)/(t)+(a)/(t^(3))` Solving, `2y=at+(2a)/(t)+(a)/(t^(3))` `rArry=(a(t^(2)+1)^(2))/(2t^(3))` |
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| 138. |
Let a,r,s,t be non-zero real numbers. Let P `(at^(2), 2at),Q R(ar^(2), 2ar)` and S `(as^(2), 2as)` be distinct point on the parabola `y^(2) = 4ax`. Suppose the PQ si the focal chord and line QR and PK are parallel, where K is point (2a, 0) It st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate isA. `((t^(2) + 1)^(2))/(2t^(3))`B. `(a(t^(2) + 1)^(2))/(2t^(3))`C. `(a(t^(2) + 1)^(2))/(t^(3))`D. `(a(t^(2) + 2)^(2))/(t^(3))` |
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Answer» Correct Answer - B Plan Equation of tangent and normal at `(at^(2), 2at)` are given by `ty = x + at^(2)` and `y + tx = 2a + at^(3)`, respectively. Tangent at `P + ty = x + at^(2)` or `y = (x)/(t) + at` Normal at `S : y (x)/(t) = (2a)/(t) + (a)/(t^(3))` Solving `2y = at + (2a)/(t) + (a)/(t^(3)) implies y = (a(t^(3) + 1)^(2))/(2t^(3))` |
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| 139. |
Let a, r, s, t be non-zero real numbers. Let `P(at^2, 2at), Q, R(ar^2, 2ar) and S(as^2, 2as)` be distinct points onthe parabola `y^2 = 4ax`. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K isthe point (2a, 0). The value of r isA. `=(1)/(t)`B. `(t^(2) + 1)/(t)`C. `(1)/(t)`D. `(t^(2) - 1)/(t)` |
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Answer» Correct Answer - D Plan (I) If `P(at^(2), 1at)` is one end point of focal chord of parabola `y^(2) = 4ax`, then other end point is `((a)/(t^(2)), (2a)/(t))` (ii) Slope of line joining two points `(x_(1), y_(1))` and `(x_(2), y_(2))` is given by `(y_(2) - y_(1))/(x_(2) - x_(1))` If PQ is focal chord, then coordinates of Q will be `((a)/(t^(2)),(2a)/(t))` Now, slope of QR = slope of PK `(2ar + (2a)/(t))/(ar^(2) - (a)/(t^(2))) = (2 at)/(at^(2) - 2a) implies (r + 1//t)/(r^(2) - 1//t^(2)) = (t)/(t^(2) - 2)` `implies (1)/(r - (1)/(t)) = (t)/(t^(2) - 2) implies r - (1)/(t) = (t^(2) - 2)/(t) = t - (2)/(t)` `implies r = t - (1)/(t) = (t^(2) - 1)/(t)` |
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| 140. |
The focal chord of the parabola `y^2=a x`is `2x-y-8=0`. Then find the equation of the directrix.A. `x+4=0`B. `X-4=0`C. `Y-4=0`D. `Y++4=0` |
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Answer» Correct Answer - A The coordinates of the focus of the parabola `y^(2)=ax` are (a/4, 0). It is given that `2x-y-8=0` is a focal chord. `:." "a/2-8=0rArra=16`. The equation of the directrix is `x=-a/4" i.e. "x=-4`. |
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| 141. |
Number of common chords of a parabola & a circle can beA. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - C A parabola and a circle can intersect in a most 4. distinct points and from these 4 points we get `""^(4)C_(2)=6` common chords. |
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| 142. |
A ray of light moving parallel to the x-axis gets reflected from parabolic mirror whose equation is `(y-3)^(2)=8(x+2)`. After reflection, the ray must pass throughA. (0, 3)B. (3, 0)C. (0, 0)D. none of these |
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Answer» Correct Answer - A We know that any ray parallel to the axis of a parabola after reflection passes through the focus of the parabola. The coordinates of the focus, are(0, 3). Hence, the ray passes through the point (0, 3). |
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| 143. |
Find the point where the line `x+y=6`is a normal to the parabola `y^2=8xdot`A. (18, -12)B. (4, 2)C. (2, 4)D. (3, 3) |
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Answer» Correct Answer - C We have, Slope of the normal= (Solpe of the line x + y = 6) = -1 The coordinates of the foot of the normal are `(am^(2), -2am)`. Here a = 2 and m = -1. So, the required point is (2, 4). |
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| 144. |
If three distinct normals are drawn from `(2k, 0)` to the parabola `y^2 = 4x` such that one of them is x-axis and other two are perpendicular, then `k =`A. 1B. `1/2`C. `3/2`D. none of these |
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Answer» Correct Answer - C The equation of any normal to `y^(2)=4x` is `y=mx-2m-m^(3)` If it passes through (2k, 0), then `m^(3)+2m(1-k)=0rArrm=0or m^(2)+(1-k)=0` Clearly, m = 0 gives y = 0 i.e., x-axis as a normal. Other two normals will be prependicular if product of the roots of `m^(2)+2(2-k)=0` is -1 i.e., `2(1-k)=-1rArrk=3/2` |
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| 145. |
If three distinct normals are drawn from `(2k, 0)` to the parabola `y^2 = 4x` such that one of them is x-axis and other two are perpendicular, then `k =`A. `klt1`B. `kgt1`C. `kle1`D. `kge1` |
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Answer» Correct Answer - B The equation of any normal to `y^(2)=4x`, is `y=mx-2m-m^(3)" …(i)"` `rArr" "m^(3)+2m(1-k)=0rArrm=0, m=+-sqrt(2(k-1))` For three normals to be real and distinct, we must have `2(k-1)gt1` |
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| 146. |
Three normals to the parabola `y^2= x` are drawn through a point `(C, O)` then C=A. `c=1//4`B. `c=1//2`C. `clt1//2`D. none of these |
| Answer» Correct Answer - C | |
| 147. |
Find the number of distinct normals that can be drawn from `(-2,1)`to the parabola `y^2-4x-2y-3=0`A. 3B. 2C. 1D. 4 |
| Answer» Correct Answer - A | |
| 148. |
At what point on the parabola `y^2=4x`the normal makes equal angle with the axes?`(4,4)`(b) `(9,6)`(d) `(4,-4)`(d) `(1,+-2)`A. (4, 4)B. (9, 6)C. (4, -4)D. (1, -2) |
| Answer» Correct Answer - D | |
| 149. |
The line `lx+my+n=0` is a normal to the parabola `y^2 = 4ax` ifA. `al(l^(2)+2m^(2))+m^(2)n=0`B. `al(l^(2)+2m^(2))+m^(2)n`C. `al(2l^(2)+2m^(2))+m^(2)n=0`D. `al(2l^(2)+2m^(2))+m^(2)n` |
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Answer» Correct Answer - A The equation of the line is lx+n=0. `or, y=((-1)/m)x+((-n)/m)` This will be normal to `y^(2)=4ax`, if `(-n)/m=-2a((-1)/m)-a((-1)/m)^(3)" "{"Using : c"=-2am-am^(3)}` `rArr" "-nm^(2)=2alm^(2)+al^(3)rArral(l^(2)+2m^(2))+m^(2)n=0` |
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| 150. |
The normal to the parabola `y^(2)=4x` at P (1, 2) meets the parabola again in Q, then coordinates of Q areA. (-6, 9)B. (9, -6)C. (-9, -6)D. (-6, -9) |
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Answer» Correct Answer - B Let the coordinates of P be `(t_(1)^(2), 2t_(1))`. But, the coordinates P are given as (1, 2). Therefore, `t_(!)=1`. Let the coordinates of Q be `(t_(2)^(2), 2t_(2))`. Then, `t_(2)=-t_(1)-2/t_(1)rArrt_(2)=-3` Hence, the coordinates of Q are (9, -6). |
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