The tangents drawn to the hyperbola 5x^2 - 9y | vedclass

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

Question

(C) The given hyperbola is $5x^2 - 9y^2 = 90$,which can be written as $\frac{x^2}{18} - \frac{y^2}{10} = 1$. Here,$a^2 = 18$ and $b^2 = 10$. The equat

Vedclass · Accepted Answer

$x^2 - y^2 = 28$

Vedclass · Answer

(C) The given hyperbola is $5x^2 - 9y^2 = 90$,which can be written as $\frac{x^2}{18} - \frac{y^2}{10} = 1$. Here,$a^2 = 18$ and $b^2 = 10$. The equation of a tangent to the hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$,which is $y = mx \pm \sqrt{18m^2 - 10}$. If this tangent passes through $P(h, k)$,then $k - mh = \pm \sqrt{18m^2 - 10}$. Squaring both sides,we get $(k - mh)^2 = 18m^2 - 10$,which simplifies to $m^2(h^2 - 18) - 2mhk + (k^2 + 10) = 0$. Let the slopes be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. Since $\alpha + \beta = 90^\circ$,we have $\beta = 90^\circ - \alpha$,so $m_2 = 	an(90^\circ - \alpha) = \cot \alpha = \frac{1}{m_1}$. Thus,$m_1 m_2 = 1$. From the quadratic equation in $m$,the product of roots is $m_1 m_2 = \frac{k^2 + 10}{h^2 - 18}$. Setting this equal to $1$,we get $k^2 + 10 = h^2 - 18$,which implies $h^2 - k^2 = 28$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 = 28$.

The tangents drawn to the hyperbola $5x^2 - 9y^2 = 90$ through a variable point $P$ make the angles $\alpha$ and $\beta$ with its transverse axis. If $\alpha$ and $\beta$ are complementary angles,then the locus of $P$ is

Similar Questions

The point of contact of the tangent $y = x + 2$ to the hyperbola $5x^2 - 9y^2 = 45$ is

If the line $4x + 3y = 7$ touches the hyperbola $x^2 - y^2 = 7$, then the sum of the coordinates of the point of contact is...

The equation of a hyperbola,whose foci are $(5, 0)$ and $(-5, 0)$ and the length of whose conjugate axis is $8$,is

If the equation of the tangent to the hyperbola $5x^2 - 9y^2 - 20x - 18y - 34 = 0$ which makes an angle of $45^{\circ}$ with the positive $X$-axis is $x + by + c = 0$,then $b^2 + c^2 =$

The length of the latus rectum of the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$ is (where $\alpha \neq \frac{n\pi}{2}, n \in I$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module