If a pair of variable straight lines $x^2 + 4y^2 + \alpha xy = 0$ (where $\alpha$ is a real parameter) cut the ellipse $x^2 + 4y^2 = 4$ at two points A and B, then the locus of the point of intersection of tangents at A and B is:
Step-by-Step Solution
Key Concept: The pair of lines x² + 4y² + αxy = 0 represents tangent lines from point P(h,k) to the ellipse x² + 4y² = 4. The homogenized form of the ellipse using the chord of contact equation xh/4 + yk = 1 must be identical to the given pair of lines, requiring coefficient matching of x², y², and xy terms.
If $P(h,k)$ is the intersection of tangents at points $A$ and $B$ on the ellipse $\frac{x^2}{4} + y^2 = 1$, the chord of contact is $\frac{xh}{4} + \frac{yk}{1} = 1$. Homogenizing the ellipse equation using this chord gives: $x^2(\frac{h^2}{16}-4) + y^2(k^2-1) + \frac{2hkxy}{4} = 0$. This must match the given pair of lines $x^2 + 4y^2 + axy = 0$. Comparing coefficients and solving yields $h^2 - 4k^2 = 0$, giving the locus $(x-2y)(x+2y) = 0$.
Correct Answer: 1,3