The angle between the tangents drawn from (1, | vedclass

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

Question

(D) The equation of a line passing through $(1, 2\sqrt{2})$ is $y - 2\sqrt{2} = m(x - 1)$,which can be written as $y = mx + (2\sqrt{2} - m)$.Since thi

Vedclass · Accepted Answer

$\pi /2$

Vedclass · Answer

(D) The equation of a line passing through $(1, 2\sqrt{2})$ is $y - 2\sqrt{2} = m(x - 1)$,which can be written as $y = mx + (2\sqrt{2} - m)$.Since this is a tangent to the hyperbola $16x^{2} - 25y^{2} = 400$,we divide by $400$ to get $\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1$.Here,$a^{2} = 25$ and $b^{2} = 16$.The condition for tangency is $c^{2} = a^{2}m^{2} - b^{2}$,where $c = 2\sqrt{2} - m$.Substituting the values: $(2\sqrt{2} - m)^{2} = 25m^{2} - 16$.$8 + m^{2} - 4\sqrt{2}m = 25m^{2} - 16$.$24m^{2} + 4\sqrt{2}m - 24 = 0$.Dividing by $4$: $6m^{2} + \sqrt{2}m - 6 = 0$.Let the roots be $m_{1}$ and $m_{2}$. The product of the slopes is $m_{1}m_{2} = \frac{-6}{6} = -1$.Since the product of the slopes is $-1$,the tangents are perpendicular to each other.Therefore,the angle between them is $\pi / 2$.

The angle between the tangents drawn from $(1, 2\sqrt{2})$ to the hyperbola $16x^{2} - 25y^{2} = 400$ is.....

Similar Questions

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

The length of the latus rectum of the hyperbola $9x^2 - 16y^2 - 18x - 32y - 151 = 0$ is

If the lengths of the transverse axis and the latus rectum of a hyperbola are $6$ and $\frac{8}{3}$ respectively,then the equation of the hyperbola is $ . . . . . . $

The locus of the midpoint of the system of parallel chords parallel to the line $y = 2x$ for the hyperbola $9x^2 - 4y^2 = 36$ is

The area of the quadrilateral formed by the foci of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and its conjugate hyperbola is (in square units):

Vedclass Test Series

Exam Paper Generator

Online Exam Module