An ellipse intersects the hyperbola 2x^2 - 2y | vedclass

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

Question

(D) Given hyperbola: $2x^2 - 2y^2 = 1 \Rightarrow \frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$.Here,$a^2 = 1/2, b^2 = 1/2$. Eccentricity $e_h = \sqrt{1 + \f

Vedclass · Accepted Answer

$(A, B)$

Vedclass · Answer

(D) Given hyperbola: $2x^2 - 2y^2 = 1 \Rightarrow \frac{x^2}{1/2} - \frac{y^2}{1/2} = 1$.Here,$a^2 = 1/2, b^2 = 1/2$. Eccentricity $e_h = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}$.Eccentricity of ellipse $e = \frac{1}{e_h} = \frac{1}{\sqrt{2}}$.Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since $e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{2}$,we have $\frac{b^2}{a^2} = \frac{1}{2} \Rightarrow a^2 = 2b^2$.Equation of ellipse: $x^2 + 2y^2 = 2b^2$.For orthogonal intersection,the product of slopes of tangents at the point of intersection $(x_0, y_0)$ is $-1$.For hyperbola: $4x - 4y y' = 0 \Rightarrow y' = \frac{x}{y}$.For ellipse: $2x + 4y y' = 0 \Rightarrow y' = -\frac{x}{2y}$.Product: $(\frac{x_0}{y_0})(-\frac{x_0}{2y_0}) = -1 \Rightarrow x_0^2 = 2y_0^2$.Substitute $x_0^2 = 2y_0^2$ into the hyperbola equation: $2(2y_0^2) - 2y_0^2 = 1$ $\Rightarrow 2y_0^2 = 1$ $\Rightarrow y_0^2 = 1/2$ and $x_0^2 = 1$.Substitute $(x_0^2, y_0^2) = (1, 1/2)$ into the ellipse equation: $1 + 2(1/2) = 2b^2$ $\Rightarrow 2b^2 = 2$ $\Rightarrow b^2 = 1$.Thus,$a^2 = 2(1) = 2$. The equation is $x^2 + 2y^2 = 2$.Foci are $(\pm ae, 0) = (\pm \sqrt{2} \cdot \frac{1}{\sqrt{2}}, 0) = (\pm 1, 0)$.Therefore,options $(A)$ and $(B)$ are correct.

Similar Questions

Let $P$ be the point of intersection of the common tangents to the parabola $y^2 = 12x$ and the hyperbola $8x^2 - y^2 = 8$. If $S$ and $S'$ denote the foci of the hyperbola where $S$ lies on the positive $x$-axis,then $P$ divides $SS'$ in the ratio:

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and the parabola $x^2 = 4(y + b)$ are such that the two foci of the ellipse and the end points of the latus rectum of the parabola are the vertices of a square. The eccentricity of the ellipse is

If $m$ is the slope of a common tangent to the curves $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ and $x^{2}+y^{2}=12$,then $12\; m^{2}$ is equal to

The tangent and normal at $P(t),$ for all real positive $t,$ to the parabola $y^2 = 4ax$ meet the axis of the parabola in $T$ and $G$ respectively. Find the angle at which the tangent at $P$ to the parabola is inclined to the tangent at $P$ to the circle passing through the points $P, T,$ and $G$.

The triangle $PQR$ of area $A$ is inscribed in the parabola $y^2 = 4ax$ such that the vertex $P$ lies at the vertex of the parabola and the base $QR$ is a focal chord. The modulus of the difference of the ordinates of the points $Q$ and $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module