Find the condition for the curves frac{x^2}{a | vedclass

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

Question

(B) Given curves are $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $(1)$ and $xy = c^2$ $(2)$.Differentiating $(1)$ with respect to $x$: $\frac{2x}{a^2} - \

Vedclass · Accepted Answer

$a^2 = b^2$

Vedclass · Answer

(B) Given curves are $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ $(1)$ and $xy = c^2$ $(2)$.Differentiating $(1)$ with respect to $x$: $\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{b^2x}{a^2y}$. Let this be $m_1$.Differentiating $(2)$ with respect to $x$: $y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$. Let this be $m_2$.For the curves to intersect orthogonally,$m_1 	imes m_2 = -1$.$\left( \frac{b^2x}{a^2y} ight) \left( -\frac{y}{x} ight) = -1$.$-\frac{b^2}{a^2} = -1 \Rightarrow a^2 = b^2$.

Find the condition for the curves $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $xy = c^2$ to intersect orthogonally.

Similar Questions

If $e_1$ and $e_2$ are respectively the eccentricities of the curves $9x^2 - 16y^2 - 144 = 0$ and $9x^2 - 16y^2 + 144 = 0$,then find the value of $\frac{e_1^2 e_2^2}{e_1^2 + e_2^2}$.

The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. The equation of the hyperbola with eccentricity $e = 2$ is

The eccentricity of the hyperbola $\frac{\sqrt{1999}}{3}(x^2 - y^2) = 1$ is

The locus of the point of intersection of lines $(x + y)t = a$ and $x - y = at$,where $t$ is the parameter,is

Let $P$ be a point on the hyperbola $x^2 - y^2 = 4$,which is at the minimum distance from $(0, -1)$. Then the distance of $P$ from the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module