Let the tangent drawn to the parabola y^2 = 2 | vedclass

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

Question

(D) The tangent to the parabola $y^2 = 24x$ at $(\alpha, \beta)$ is given by $y\beta = 12(x + \alpha)$. The slope of this tangent is $m_1 = \frac{12}{

Vedclass · Accepted Answer

$(15, 13)$

Vedclass · Answer

(D) The tangent to the parabola $y^2 = 24x$ at $(\alpha, \beta)$ is given by $y\beta = 12(x + \alpha)$. The slope of this tangent is $m_1 = \frac{12}{\beta}$.Given the line $2x + 2y = 5$,its slope is $m_2 = -1$. Since the tangent is perpendicular to this line,$m_1 	imes m_2 = -1$,so $\frac{12}{\beta} 	imes (-1) = -1$,which gives $\beta = 12$.Since $(\alpha, \beta)$ lies on $y^2 = 24x$,we have $12^2 = 24\alpha$,so $144 = 24\alpha$,which gives $\alpha = 6$.The hyperbola is $\frac{x^2}{36} - \frac{y^2}{144} = 1$. The point on the hyperbola is $(\alpha + 4, \beta + 4) = (10, 16)$.The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at $(x_0, y_0)$ is $\frac{a^2x}{x_0} + \frac{b^2y}{y_0} = a^2 + b^2$.Here $a^2 = 36, b^2 = 144, x_0 = 10, y_0 = 16$. The normal is $\frac{36x}{10} + \frac{144y}{16} = 36 + 144$,which simplifies to $3.6x + 9y = 180$,or $2x + 5y = 100$.Checking the options: For $(25, 10)$,$2(25) + 5(10) = 50 + 50 = 100$ (Passes).For $(20, 12)$,$2(20) + 5(12) = 40 + 60 = 100$ (Passes).For $(30, 8)$,$2(30) + 5(8) = 60 + 40 = 100$ (Passes).For $(15, 13)$,$2(15) + 5(13) = 30 + 65 = 95 
eq 100$ (Does not pass).

Let the tangent drawn to the parabola $y^2 = 24x$ at the point $(\alpha, \beta)$ be perpendicular to the line $2x + 2y = 5$. Then the normal to the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ at the point $(\alpha + 4, \beta + 4)$ does $NOT$ pass through which of the following points?

Similar Questions

If the angle between the asymptotes of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \tan^{-1}\left(\frac{b}{a}\right) = 2 \tan^{-1}\left(\frac{2}{3}\right)$ and $a^2-b^2=45$,then $ab=$

The equation $13[{(x - 1)^2} + {(y - 2)^2}] = 3{(2x + 3y - 2)^2}$ represents

If a hyperbola has a transverse axis of length $2 \sin \theta$ and is confocal with the ellipse $3x^2 + 4y^2 = 12$,then its equation is:

If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$,then the coordinates of the point of contact are

$P(\theta)$ is a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{9} = 1$,$S$ is its focus lying on the positive $X$-axis and $Q = (0, 1)$. If $S Q = \sqrt{26}$ and $S P = 6$,then $\theta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module