The line y=x+1 meets the ellipse x^24+y^22=1 at two points P and Q.

Question

The line $y=x+1$ meets the ellipse $\dfrac{x^2}4+\dfrac{y^2}2=1$ at two points $P$ and $Q$. If $r$ is the radius of the circle with $PQ$ as diameter, then $(3r)^2$ is equal to:

Accepted Answer

Let point (a, a + 1) as the point of intersection of line and ellipse.

So, $${{{a^2}} \over 4} + {{{{(a + 1)}^2}} \over 2} = 1 \Rightarrow {a^2} + 2({a^2} + 2a + 1) = 4$$

$$ \Rightarrow 3{a^2} + 4a - 2 = 0$$

If roots of this equation are $$\alpha$$ and $$\beta$$.

So, $$P(\alpha ,\,\alpha  + 1)$$ and $$Q(\beta ,\,\beta  + 1)$$

$$PQ = 4{r^2} = {(\alpha  - \beta )^2} + {(\alpha  - \beta )^2}$$

$$ \Rightarrow 9{r^2} = {9 \over 4}(2{(\alpha  - \beta )^2})$$

$$ = {9 \over 2}\left[ {{{(\alpha  + \beta )}^2} - 4\alpha \beta } \right]$$

$$ = {9 \over 2}\left[ {{{\left( { - {4 \over 3}} \right)}^2} + {8 \over 3}} \right]$$

$$ = {1 \over 2}[16 + 24] = 20$$

Solutions · 1