The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$
The point $(4, -3)$ with respect to the ellipse $4{x^2} + 5{y^2} = 1$
- A
Lies on the curve
- B
Is inside the curve
- C
Is outside the curve
- D
Is focus of the curve
Similar Questions
Let the ellipse, $E _1: \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1, a > b$ and $E _2: \frac{ x ^2}{A^2}+\frac{ y ^2}{B^2}=1, A< B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $A B C D$ equals:
- [JEE MAIN 2025]
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