Question

The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$.If $M$ is the mid-point of the line segment $PQ$, then the locus of $M$ intersects the latus rectum of the given ellipse at the points

• A. $(\pm \frac{3\sqrt{5}}{2},\pm \frac{2}{7})$
• B. $(\pm \frac{3\sqrt{5}}{2},\pm \frac{\sqrt{19}}{4})$
• C. $(\pm 2\sqrt{3},\pm \frac{1}{7})$
• D. $(\pm 2\sqrt{3},\pm \frac{4\sqrt{3}}{7})$

Answer 1

The correct option is C $(\pm 2\sqrt{3},\pm \frac{1}{7})$

$EquationofnormalatpointPis:\frac{ax}{cos\theta }-\frac{by}{sin\theta }={\left(ae\right)}^{2}X-intercept:X=\frac{{\left(ae\right)}^2\ast cos\theta }{a}LetM(h,k)bethemid-pointofPQthenh=\frac{acos\theta (1+e^2)}{2}andk=\frac{bsin\theta }{2}Or,cos\theta =\frac{2h}{a(1+e^2)}andsin\theta =\frac{2k}{b}Squaringandaddingweget:\frac{4h^2}{{\left(a\right(1+e^2\left)\right)}^2}+\frac{{4k}^2}{b^2}=1a=4b=2e=\frac{\sqrt{3}}{2}\therefore \frac{4h^2}{49}+\frac{k^2}{1}=1----------\left(1\right)Equationoflatusrectumisx=\pm ae=\pm 2\sqrt{3}Pointofintersectionwithequation1is:\frac{4\ast 12}{49}+\frac{k^2}{1}=1or,k^2=\frac{1}{49}k=\pm \frac{1}{7}\therefore pointsare(\pm 2\sqrt{3},\pm \frac{1}{7})$

Option [C]