Angle between Two Straight Lines

Q. Let $E_1$ and $E_2$ be two ellipses whose centers are in origin. The major axes of $E_1$ and $E_2$ lies along the $x-$axis and $y-$axis, respectively. Let $S$ be the circle $x^2+(y-1{)}^2=2.$. The straight lin $x+y=3$ touches the curves $S$, $E_1$ and $E_2$ at $P,Q$ and $R,$ respectively. Supoose the $PQ=PR=\frac{2\sqrt{2}}{3}.$If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression(s) is(are)

1. $e_1^2+e_2^2=\frac{43}{40}$
2. $e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{1}0}$
3. $|e_1^2-e_2^2|=\frac{5}{8}$
4. $e_1{e}_2=\frac{\sqrt{3}}{4}$

Q. ​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(4)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the CORRECT option is :

1. $\left(4\right)\to \left(S\right)$
2. $\left(3\right)\to \left(P\right)$
3. $\left(2\right)\to \left(Q\right)$
4. $\left(1\right)\to \left(T\right)$

Q. A pair of straight lines through $A(2,7)$ is drawn to intersect the line $x+y=5$ at $C$ and $D.$ If angle between the pair of straight lines is $\frac{\pi }{2},$ then the locus of incentre of $△ACD$ is

1. $2xy-6x+4y=28$
2. $xy-6x+4y=29$
3. $x^2+y^2-6x+4y=29$
4. $xy-6x+4y=28$

Q. ​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the INCORRECT option is:

1. $\left(1\right)\to \left(R\right)$
2. $\left(2\right)\to \left(R\right)$
3. $\left(3\right)\to \left(S\right)$
4. $\left(4\right)\to \left(S\right)$

Q. ​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the INCORRECT option is:

1. $\left(1\right)\to \left(R\right)$
2. $\left(2\right)\to \left(R\right)$
3. $\left(3\right)\to \left(S\right)$
4. $\left(4\right)\to \left(S\right)$

Q. ​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the INCORRECT option is:

1. $\left(1\right)\to \left(R\right)$
2. $\left(2\right)\to \left(R\right)$
3. $\left(3\right)\to \left(S\right)$
4. $\left(4\right)\to \left(S\right)$

Q. ​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(4)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the CORRECT option is :

1. $\left(1\right)\to \left(T\right)$
2. $\left(2\right)\to \left(Q\right)$
3. $\left(3\right)\to \left(P\right)$
4. $\left(4\right)\to \left(S\right)$

Q. ​​​​​​A circle $C_1$ with centre at the origin meets $x$-axis at $A$ and $B$ (where $A\&B$ lies on negative and positive $x-$axis respectively). Two points $P\left(a\right)$ and $Q\left(b\right)$ are on the circle such that $b-a$ is a constant, where $a$ and $b$ are the parametric angles of the points. $BP$ and $AQ$ meets at $R$. Locus of $R$ is a circle $C_2$.
Let $c$ be the radius of $C_1$ and $d$ be the radius of $C_2$.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \text{For}b-a=\frac{\pi }{2}\text{, the value of}\frac{d^2}{c^2}\text{is}& \text{(P)}& 0\\ \text{(2)}& \text{For}b-a=\frac{\pi }{2}\text{and}c=\sqrt{2}\text{, circle}{C}_2\text{intersects}& \text{(Q)}& 1\\ & \text{the coordinate axes at four points}L,M,N,O.& & \\ & \text{Let the area of the quadrilateral LMNO is}2\sqrt{2}p.& & \\ & \text{Then the value of}p\text{is}& & \\ \text{(3)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(R\right)& 2\\ & \text{respectively.}\text{If}{m}_1{m}_2=-1\text{, then}\frac{a}{b}\text{is}& & \\ \text{(4)}& \text{Let}{m}_1,m_2\text{be the slopes of the line BQ, AP}& \left(S\right)& 3\\ & \text{respectively.}\text{If}{m}_1=m_2\text{, then}\frac{3|b-a|}{\pi }\text{is}& & \\ & & \left(T\right)& 4\end{array}$

Then the CORRECT option is :

1. $\left(1\right)\to \left(T\right)$
2. $\left(2\right)\to \left(Q\right)$
3. $\left(3\right)\to \left(P\right)$
4. $\left(4\right)\to \left(S\right)$

Q.

Shape	Centre of rotation	Order of rotation	angle of rotation
Equilateral Triangle	P
Rectangle		Q
Square	R		S

Find P, Q, R and S respectively.

1. Intersection point of medians, 2, intersection point of diagonals, ${90}^o$
2. Intersection point of medians, 2, intersection point of diagonals, ${180}^o$
3. Intersection point of sides, 4, intersection point of diagonals, ${45}^o$
4. Intersection point of altitudes, 4, intersection point of diagonals, ${25}^o$