If the line 2x + sqrt{6}y = 2 touches the hyp | vedclass

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

Question

(B) The given line is $2x + \sqrt{6}y = 2$,which can be written as $2x + \sqrt{6}y - 2 = 0$.The equation of the hyperbola is $x^2 - 2y^2 = 4$.Let the

Vedclass · Accepted Answer

$(4, -\sqrt{6})$

Vedclass · Answer

(B) The given line is $2x + \sqrt{6}y = 2$,which can be written as $2x + \sqrt{6}y - 2 = 0$.The equation of the hyperbola is $x^2 - 2y^2 = 4$.Let the point of contact be $(x_1, y_1)$.The equation of the tangent to the hyperbola $x^2 - 2y^2 = 4$ at $(x_1, y_1)$ is given by $xx_1 - 2yy_1 = 4$,or $xx_1 - 2yy_1 - 4 = 0$.Since this line is the same as the given line $2x + \sqrt{6}y - 2 = 0$,the coefficients must be proportional:$\frac{x_1}{2} = \frac{-2y_1}{\sqrt{6}} = \frac{-4}{-2}$$\frac{x_1}{2} = 2 \Rightarrow x_1 = 4$$\frac{-2y_1}{\sqrt{6}} = 2$ $\Rightarrow -2y_1 = 2\sqrt{6}$ $\Rightarrow y_1 = -\sqrt{6}$Thus,the point of contact is $(4, -\sqrt{6})$.

If the line $2x + \sqrt{6}y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$,then the coordinates of the point of contact are

Similar Questions

From any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,tangents are drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 2$. The area of the figure formed by the chord of contact of that point and the asymptotes is

What kind of hyperbola does the equation $9x^2 - 16y^2 - 18x + 32y - 151 = 0$ represent?

The asymptotes of a hyperbola are parallel to $2x + 3y = 0$ and $3x + 2y = 0$. The equation of the hyperbola whose center is at $(1, 2)$ and which passes through $(5, 3)$ is:

The equation of the hyperbola whose coordinates of the foci are $(\pm 8, 0)$ and the length of the latus rectum is $24$ units,is

Find the length of the latus rectum of the hyperbola $16x^2 - 9y^2 = 144$. (in $/3$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module