The point of contact of the tangent y = x + 2 | vedclass

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

Question

(C) The given hyperbola is $5x^2 - 9y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} - \frac{y^2}{5} = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^

Vedclass · Accepted Answer

$(-9/2, -5/2)$

Vedclass · Answer

(C) The given hyperbola is $5x^2 - 9y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} - \frac{y^2}{5} = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 5$. The tangent line is $y = mx + c$,where $m = 1$ and $c = 2$. The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left( \frac{-a^2m}{c}, \frac{-b^2}{c} \right)$. Substituting the values: $x_1 = \frac{-9(1)}{2} = -\frac{9}{2}$ and $y_1 = \frac{-5}{2}$. Thus,the point of contact is $(-9/2, -5/2)$.

The point of contact of the tangent $y = x + 2$ to the hyperbola $5x^2 - 9y^2 = 45$ is

Similar Questions

For the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 1$,which of the following remains constant with a change in $\alpha$?

If $5x^2 + \lambda y^2 = 20$ represents a rectangular hyperbola,then $\lambda$ equals:

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

If $\theta$ is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity $e = \frac{2}{\sqrt{7}-\sqrt{3}}$,then $\sin \theta = $

Let $S$ be the focus of the hyperbola $x^2 - 2y^2 = 1$ lying on the positive $X$-axis. Let $P(-1, 1)$ be a given point. Then the area of the triangle formed by the line $PS$ with the coordinate axes is (in sq. units)

Vedclass Test Series

Exam Paper Generator

Online Exam Module