- Home
- Standard 11
- Mathematics
$\text { Let } $S$ \text { be the circle in the } xy \text {-plane defined by the equation } x ^2+ y ^2=4 \text {. }$
($1$) Let $E_1, E_2$ and $F_1 F_2$ be the chords of $S$ passing through the point $P_0(1,1)$ and parallel to the $x$-axis and the $y$-axis, respectively. Let $G _1 G _2$ be the chord of $S$ passing through $P _0$ and having slope -$1$ . Let the tangents to $S$ at $E_1$ and $E_2$ meet at $E_3$, the tangents to $S$ at $F_1$ and $F_2$ meet at $F_3$, and the tangents to $S$ at $G_1$ and $G_2$ meet at $G_3$. Then, the points $E_3, F_3$, and $G _3$ lie on the curve
$(A)$ $x+y=4$ $(B)$ $(x-4)^2+(y-4)^2=16$ $(C)$ $(x-4)(y-4)=4$ $(D)$ $x y=4$
($2$) Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$. Then, the mid-point of the line segment MN must lie on the curve
$(A)$ $(x+y)^2=3 x y$ $(B)$ $x^{2 / 3}+y^{2 / 3}=2^{4 / 3}$ $(C)$ $x^2+y^2=2 x y$ $(D)$ $x^2+y^2=x^2 y^2$
Give the answer or quetion ($1$) and ($2$)
$A,D$
$A,B$
$A,C$
$A,B,C$
Solution
(Image)
($1$) $E_1(-\sqrt{3}, 1), E_2(\sqrt{3}, 1), F_1(1, \sqrt{3}), F_2(1,-\sqrt{3}), G_1(0,2), G_2(2,0)$
Tangent at $\quad E_2(\sqrt{3}, 1)$ is $\quad x \sqrt{3}+ y =4$
$E _1(-\sqrt{3}, 1) \text { is } \quad- x \sqrt{3}+ y =4$
Solving $(0,4)=E_3$
Similarly $(4,0)= F _3$
$(2,2)=G_3$
$E _3, F _3, G _3$ satisfy option $(A)$
($2$) Tangent $x \cos \theta+y \sin \theta=2$
$M \left(\frac{2}{\cos \theta}, 0\right), N \left(0, \frac{2}{\sin \theta}\right)$Let $P ( h , k )$ mid point of $M$ and $N$
$P ( h , k )=\left(\frac{\frac{2}{\cos \theta}+0}{2}, \frac{0+\frac{2}{\sin \theta}}{2}\right)$
$\therefore \cos \theta=\frac{1}{ h }, \quad \sin \theta=\frac{1}{ k }$
$\cos ^2 \theta+\sin ^2 \theta=1$
$\frac{1}{ h ^2}+\frac{1}{ k ^2}=1$
$h ^2+ k ^2= h ^2 k ^2$
Locus $x ^2+ y ^2= x ^2 y ^2$
