A rod of length $12 \,cm$ moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is $3\, cm$ from the end in contact with the $x-$ axis.
Let $AB$ be the rod making an angle $\theta$ with $O X$ and $P ( x ,\, y )$ be the point on it such that $AP =3\,cm$
Then, $PB = AB - AP =(12-3)\, cm =9\, cm$ $[ AB =12 \,cm ]$
From $P$, draw $PQ \perp OY$ and $PR \perp OX$.
In $\Delta PBQ$ , $\cos \theta=\frac{ PQ }{ PB }=\frac{x}{9}$
In $\Delta PRA$ , $\sin \theta=\frac{ PR }{ PA }=\frac{y}{3}$
since, $\sin ^{2} \theta+\cos ^{2} \theta=1$
$\left(\frac{y}{3}\right)^{2}+\left(\frac{x}{9}\right)^{2}=1$
Or, $\frac{x^{2}}{81}+\frac{y^{2}}{9}=1$
Thus, the equation of the locus of point $P$ on the rod is $\frac{x^{2}}{81}+\frac{y^2} {9}=1$.
The eccentricity of an ellipse whose centre is at the origin is $\frac{1}{2}$ . If one of its directices is $x = - 4$ then the equation of the normal to it at $\left( {1,\frac{3}{2}} \right)$ is
Area of the quadrilaterals formed by drawing tangents at the ends of latus recta of $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1$ is
Let $F_1\left(x_1, 0\right)$ and $F_2\left(x_2, 0\right)$, for $x_1<0$ and $x_2>0$, be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$. Suppose a parabola having vertex at the origin and focus at $F_2$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
($1$)The orthocentre of the triangle $F_1 M N$ is
($A$) $\left(-\frac{9}{10}, 0\right)$ ($B$) $\left(\frac{2}{3}, 0\right)$ ($C$) $\left(\frac{9}{10}, 0\right)$ ($D$) $\left(\frac{2}{3}, \sqrt{6}\right)$
($2$) If the tangents to the ellipse at $M$ and $N$ meet at $R$ and the normal to the parabola at $M$ meets the $x$-axis at $Q$, then the ratio of area of the triangle $M Q R$ to area of the quadrilateral $M F_{\mathrm{I}} N F_2$ is
($A$) $3: 4$ ($B$) $4: 5$ ($C$) $5: 8$ ($D$) $2: 3$
Givan the answer qestion ($1$) and ($2$)
The equation to the locus of the middle point of the portion of the tangent to the ellipse $\frac{{{x^2}}}{{16}}$$+$ $\frac{{{y^2}}}{9}$ $= 1$ included between the co-ordinate axes is the curve :
The eccentricity of the ellipse $25{x^2} + 16{y^2} - 150x - 175 = 0$ is