The straight line x + y = sqrt{2}p will touch | vedclass

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

Question

(D) The given hyperbola is $4x^2 - 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The line is

Vedclass · Accepted Answer

$2p^2 = 5$

Vedclass · Answer

(D) The given hyperbola is $4x^2 - 9y^2 = 36$. Dividing by $36$,we get $\frac{x^2}{9} - \frac{y^2}{4} = 1$. Here,$a^2 = 9$ and $b^2 = 4$. The line is $x + y = \sqrt{2}p$,which can be written as $y = -x + \sqrt{2}p$. Comparing this with $y = mx + c$,we have $m = -1$ and $c = \sqrt{2}p$. The condition for the line $y = mx + c$ to touch the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$. Substituting the values,we get $(\sqrt{2}p)^2 = 9(-1)^2 - 4$. $2p^2 = 9 - 4$. $2p^2 = 5$.

The straight line $x + y = \sqrt{2}p$ will touch the hyperbola $4x^2 - 9y^2 = 36$,if

Similar Questions

If $x = 9$ is the chord of contact of the hyperbola $x^2 - y^2 = 9$,then the equation of the corresponding pair of tangents is

The eccentricity of the hyperbola $2x^2 - y^2 = 6$ is

Let the foci of a hyperbola be $(1, 14)$ and $(1, -12)$. If it passes through the point $(1, 6)$,then the length of its latus-rectum is:

The equation of a tangent to the hyperbola $5x^{2}-y^{2}=5$ which passes through the external point $(2, 8)$ is:

$A$ hyperbola has its transverse axis along the major axis of the conic $\frac{x^2}{3} + \frac{y^2}{4} = 4$ and its vertices at the foci of this conic. If the eccentricity of the hyperbola is $\frac{3}{2}$,then which of the following points does $NOT$ lie on it?

Vedclass Test Series

Exam Paper Generator

Online Exam Module