The locus of middle points of the chords of t | vedclass

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

Question

Equation of chord of the given circle whose middle point is

$(\mathrm{h}, \mathrm{k})$ is $\mathrm{xh}+\mathrm{yk}=\mathrm{h}^{2}+\mathrm{k}^{2}$

Vedclass · Accepted Answer

${({x^2} + {y^2})^2} = {a^2}{x^2} - {b^2}{y^2}$

Vedclass · Answer

Equation of chord of the given circle whose middle point is

$(\mathrm{h}, \mathrm{k})$ is $\mathrm{xh}+\mathrm{yk}=\mathrm{h}^{2}+\mathrm{k}^{2}$

and equation of tangent line

upon hyperbola is $\mathrm{y}-\mathrm{mx}=\sqrt{\mathrm{a}^{2} \mathrm{m}^{2}-\mathrm{b}^{2}}$

As both lines are identical

$\Rightarrow-\frac{m}{h}=\frac{1}{k}=\frac{\sqrt{a^{2} m^{2}-b^{2}}}{h^{2}+k^{2}}$

Eliminating $m$ 

$\Rightarrow\left(\mathrm{h}^{2}+\mathrm{k}^{2}\right)^{2}=\left(\mathrm{a}^{2} \mathrm{h}^{2}-\mathrm{b}^{2} \mathrm{k}^{2}\right)$

hence the locus is $\left(x^{2}+y^{2}\right)^{2}=\left(a^{2} x^{2}-b^{2} y^{2}\right)$

The locus of middle points of the chords of the circle $x^2 + y^2 = a^2$ which touch the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is

Similar Questions

Find the equation of the hyperbola satisfying the give conditions: Vertices $(0,\,\pm 5),$ foci $(0,\,±8)$

Eccentricity of the curve ${x^2} - {y^2} = {a^2}$ is

The foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and the hyperbola $\frac{{{x^2}}}{{144}} - \frac{{{y^2}}}{{81}} = \frac{1}{{25}}$ coincide. Then the value of $b^2$ is -

The equation of the normal to the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$ at $( - 4,\;0)$ is

Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 3 \sqrt{5},\,0),$ the latus rectum is of length $8$