1

If a pair of variable straight lines $x^2+4y^2+\alpha xy=0$ (where α is a real parameter) cuts the ellipse $x^2+4y^2=4$ at two points A and B, then the locus of the point intersection of tangents at A and B is…

${x^2} + 4{y^2} + \alpha xy = 0$

${x^2} + 4{y^2} = 4$

$ - \alpha xy = 4 \Rightarrow xy = - \frac{4}{\alpha } \Rightarrow x = \frac{2}{{\sqrt \alpha }};y = - \frac{2}{{\sqrt \alpha }}$

Hence there are two curves $x^2+4y^2=4$and $xy = - \frac{4}{\alpha }$

The parametric points are $x = \frac{2}{{\sqrt \alpha }};y = - \frac{2}{{\sqrt \alpha }}$ considering the region in the FOURTH quadrant

How do we proceed from here

Samar Imam Zaidi
  • 8,800
  • 2
    Replace $u=2y$ to transform the ellipse to a circle. Because of symmetry you can show that all intersection points are at $u=\pm x$. Transforming back you get $y = \pm x/2$. – blamocur Nov 20 '22 at 14:17

1 Answers1

1

If a pair of variable straight lines $x^2+4y^2+\alpha xy=0$ (where α is a real parameter) cuts the ellipse $x^2+4y^2=4$ at two points A and B, then the locus of the point intersection of tangents at A and B is?

Using the fact that a pair of straight lines can be "factored" into two individual straight lines.

The given pair of straight lines is $x^2+4y^2+\alpha xy=0$;

$$x^2+4y^2+\alpha xy=(x-ky)(x-ly)=x^2-xy(l+k)+kly^2=0$$

On comparing, we find; $kl=4\tag1$

Using (1), we get the two individual straight lines as;

$$x-ky=0\tag2$$

$$x-\frac{4}{k}y=0\tag3$$

The given conic(ellipse) is $x^2+4y^2=4\tag4$

Let on solving (2) and (4) give point A

Let on solving (3) and (4) give point B;

$$A\left(\frac{2k}{\sqrt{k^2+4}},\frac{2}{\sqrt{k^2+4}}\right)$$

$$B\left(\frac{4}{\sqrt{k^2+4}},\frac{k}{\sqrt{k^2+4}}\right)$$

To get the tangent equation at any point on the ellipse, we can differentiate it to get the slope at that point and use point-slope form. Following this, we get two tangent equations at A and B respectively. I think you can take it from here.

NadiKeUssPar
  • 2,474