Aspire Faculty ID #16711 · Topic: CUET 2023 · Just now
CUET 2023

The tangent to the hyperbola $x^2-y^2=3$ are parallel to the straight line $2x+y+8=0$ at the following points:

Solution

Slope of line $2x+y+8=0$ is $-2$. Slope of tangent to hyperbola is $\dfrac{x}{y}$. Set $\dfrac{x}{y}=-2 \Rightarrow y=-\dfrac{x}{2}$. Substitute in $x^2-y^2=3$ ⇒ $x=\pm2$, $y=\mp1$.

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