Question

A straight line through A(6, 8) meets the curve $2x^2+y^2=2$ at B and C. P is a point on BC such that AB, AP, AC are in H.P, then the minimum distance of the origin from the locus of ‘P’ is

• A. $\frac{67}{\sqrt{52}}$
• B. $\frac{5}{\sqrt{52}}$
• C. $\frac{10}{\sqrt{52}}$
• D. $\frac{15}{\sqrt{52}}$

Answer 1

The correct option is A $\frac{67}{\sqrt{52}}$

$(6+rcos\theta ,8+rsin\theta )$ lies on $2x^2+y^2=2$
$\Rightarrow (2co{s}^2\theta +si{n}^2\theta )r^2+2(12cos\theta +8sin\theta )r+134=0$
AB, AP, AC are in H.P $\Rightarrow \frac{2}{r}=\frac{AB+AC}{AB.AC}\Rightarrow \frac{1}{r}=-\frac{(6cos\theta +4sin\theta )}{67}\Rightarrow 6x+4y-1=0$
Minimum distance from 'O' $=\frac{1}{\sqrt{52}}$.