Let P(alpha, beta) be a point on the parabola | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the chord of the parabola $x^2 = 8y$ with midpoint $(x_1, y_1) = (1, 5/4)$ is given by $T = S_1$.Here,$T = x x_1 - 4(y + y_1)$ and

Vedclass · Accepted Answer

$192$

Vedclass · Answer

(C) The equation of the chord of the parabola $x^2 = 8y$ with midpoint $(x_1, y_1) = (1, 5/4)$ is given by $T = S_1$.Here,$T = x x_1 - 4(y + y_1)$ and $S_1 = x_1^2 - 8y_1$.Substituting the values,we get $x(1) - 4(y + 5/4) = 1^2 - 8(5/4)$.$x - 4y - 5 = 1 - 10$.$x - 4y - 5 = -9 \Rightarrow x - 4y + 4 = 0$.Since $P(\alpha, \beta)$ lies on this chord,$\alpha - 4\beta + 4 = 0$,which implies $\alpha = 4\beta - 4$.Also,$P(\alpha, \beta)$ lies on $y^2 = 4x$,so $\beta^2 = 4\alpha$.Substituting $\alpha$ in the second equation: $\beta^2 = 4(4\beta - 4) = 16\beta - 16$.$\beta^2 - 16\beta + 16 = 0$.This gives $\beta = \frac{16 \pm \sqrt{256 - 64}}{2} = 8 \pm \sqrt{48} = 8 \pm 4\sqrt{3}$.Since $\alpha = 4\beta - 4$,we have $\alpha - 28 = 4\beta - 4 - 28 = 4\beta - 32 = 4(\beta - 8)$.Thus,$(\alpha - 28)(\beta - 8) = 4(\beta - 8)^2$.Since $(\beta - 8) = \pm 4\sqrt{3}$,then $(\beta - 8)^2 = 48$.Therefore,$(\alpha - 28)(\beta - 8) = 4 	imes 48 = 192$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$. If $P$ also lies on the chord of the parabola $x^2 = 8y$ whose midpoint is $(1, 5/4)$,then $(\alpha - 28)(\beta - 8)$ is equal to:

Similar Questions

Two parabolas with a common vertex at the origin and with axes along the $x-$ axis and $y-$ axis,respectively,intersect each other in the first quadrant. If the length of the latus rectum of each parabola is $3$,then the equation of the common tangent to the two parabolas is?

The length of the latus rectum of the conic $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ is

The line $y=x+1$ is a tangent to the curve $y^{2}=4x$ at the point

If the line $2bx + 3cy + 4d = 0$ passes through the points of intersection of $y^2 = 4ax$ and $x^2 = 4ay$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module