The line y=x intersects the hyperbola frac{x^ | vedclass

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

Question

(D) Given the equations of the hyperbola and the line are $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ and $y=x$ respectively.To find the intersection points,

Vedclass · Accepted Answer

$\frac{2 \sqrt{2}}{3}$

Vedclass · Answer

(D) Given the equations of the hyperbola and the line are $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ and $y=x$ respectively.To find the intersection points, substitute $y=x$ into the hyperbola equation:$\frac{x^{2}}{9}-\frac{x^{2}}{25}=1$$\frac{25x^{2}-9x^{2}}{225}=1$ $\Rightarrow 16x^{2}=225$ $\Rightarrow x^{2}=\frac{225}{16}$$x=\pm \frac{15}{4}$. Since $y=x$, the intersection points are $P\left(\frac{15}{4}, \frac{15}{4}\right)$ and $Q\left(-\frac{15}{4}, -\frac{15}{4}\right)$.The length of the segment $PQ$ (major axis) is the distance between $P$ and $Q$:$2a = \sqrt{(\frac{15}{4} - (-\frac{15}{4}))^{2} + (\frac{15}{4} - (-\frac{15}{4}))^{2}} = \sqrt{(\frac{30}{4})^{2} + (\frac{30}{4})^{2}} = \sqrt{2 \cdot (\frac{15}{2})^{2}} = \frac{15}{\sqrt{2}}$.Thus, $a = \frac{15}{2\sqrt{2}}$.The length of the minor axis is $2b = \frac{5}{\sqrt{2}}$, so $b = \frac{5}{2\sqrt{2}}$.The eccentricity $e$ of the ellipse is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$.$e = \sqrt{1 - \frac{(5/2\sqrt{2})^{2}}{(15/2\sqrt{2})^{2}}} = \sqrt{1 - (\frac{5}{15})^{2}} = \sqrt{1 - (\frac{1}{3})^{2}} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.

The line $y=x$ intersects the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ at the points $P$ and $Q$. The eccentricity of the ellipse with $PQ$ as the major axis and a minor axis of length $\frac{5}{\sqrt{2}}$ is

Similar Questions

Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its foci at $P(-3,0)$. If the latus rectum through its other focus subtends a right angle at $P$ and $a^2b^2 = \alpha\sqrt{2} - \beta$,where $\alpha, \beta \in N$,then find the value of $\alpha + \beta$.

What is the area of the triangle formed by the lines $x - y = 0$,$x + y = 0$,and any tangent to the hyperbola $x^{2} - y^{2} = a^{2}$?

The eccentricity of the conjugate hyperbola of $16x^2 - 9y^2 - 32x - 36y - 164 = 0$ is:

Let $S$ be the focus of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ lying on the positive $X$-axis and $P(5, y_1)$ be a point on the hyperbola. Then $SP =$

If $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$ is a hyperbola,then which of the following statements can be true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module