If two points P and Q on the hyperbola frac{x | vedclass

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

Question

(B) Let the equation of line $CP$ be $y = mx$. Substituting this into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $\frac{x^2

Vedclass · Accepted Answer

$\frac{1}{a^2} - \frac{1}{b^2}$

Vedclass · Answer

(B) Let the equation of line $CP$ be $y = mx$. Substituting this into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $\frac{x^2}{a^2} - \frac{m^2 x^2}{b^2} = 1$.Solving for $x^2$,we have $x^2 = \frac{a^2 b^2}{b^2 - a^2 m^2}$.Since $y^2 = m^2 x^2$,we have $y^2 = \frac{a^2 b^2 m^2}{b^2 - a^2 m^2}$.Then $(CP)^2 = x^2 + y^2 = x^2(1 + m^2) = \frac{a^2 b^2(1 + m^2)}{b^2 - a^2 m^2}$.Since $CP \perp CQ$,the slope of $CQ$ is $-\frac{1}{m}$. Replacing $m$ with $-\frac{1}{m}$ in the expression for $(CP)^2$,we get $(CQ)^2 = \frac{a^2 b^2(1 + (-1/m)^2)}{b^2 - a^2(-1/m)^2} = \frac{a^2 b^2(1 + m^2)/m^2}{(b^2 m^2 - a^2)/m^2} = \frac{a^2 b^2(1 + m^2)}{b^2 m^2 - a^2}$.Now,$\frac{1}{(CP)^2} + \frac{1}{(CQ)^2} = \frac{b^2 - a^2 m^2}{a^2 b^2(1 + m^2)} + \frac{b^2 m^2 - a^2}{a^2 b^2(1 + m^2)} = \frac{b^2 - a^2 m^2 + b^2 m^2 - a^2}{a^2 b^2(1 + m^2)} = \frac{b^2(1 + m^2) - a^2(1 + m^2)}{a^2 b^2(1 + m^2)} = \frac{b^2 - a^2}{a^2 b^2} = \frac{1}{a^2} - \frac{1}{b^2}$.

If two points $P$ and $Q$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with centre $C$ are such that $CP$ is perpendicular to $CQ$,where $a < b$,then the value of $\frac{1}{(CP)^2} + \frac{1}{(CQ)^2}$ is:

Similar Questions

The foci of the hyperbola $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$ are:

If $P(x_1, y_1)$ is a point on the hyperbola $x^2 - y^2 = a^2$,then $SP \cdot S'P = \_\_\_\_$

The locus of the mid-point of the chord of the hyperbola $3x^2 - 2y^2 + 4x - 6y = 0$ which is parallel to $y = 2x$ is:

Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola $a^{2}x^{2} - y^{2} = b^{2}$. Then $\left(\frac{\lambda}{a}\right)^{2} - \left(\frac{\mu}{b}\right)^{2}$ is equal to

The locus of the point of intersection of the lines $\sqrt{3}x - y - 4\sqrt{3}k = 0$ and $\sqrt{3}kx + ky - 4\sqrt{3} = 0$ for different real values of $k$ is a hyperbola $H$. If $e$ is the eccentricity of $H$,then $4e^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module