The line through P(a, 2),where a neq 0,making | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The equation of the line passing through $P(a, 2)$ with an angle of $45^{\circ}$ to the $X$-axis is given by $\frac{x-a}{\cos 45^{\circ}} = \frac{

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) The equation of the line passing through $P(a, 2)$ with an angle of $45^{\circ}$ to the $X$-axis is given by $\frac{x-a}{\cos 45^{\circ}} = \frac{y-2}{\sin 45^{\circ}} = r$,which simplifies to $x = a + \frac{r}{\sqrt{2}}$ and $y = 2 + \frac{r}{\sqrt{2}}$.For point $B$ on the $X$-axis,$y=0 \Rightarrow 2 + \frac{r}{\sqrt{2}} = 0 \Rightarrow r = -2\sqrt{2}$. Thus $PB = |r| = 2\sqrt{2}$.For point $C$ on the $Y$-axis,$x=0 \Rightarrow a + \frac{r}{\sqrt{2}} = 0 \Rightarrow r = -a\sqrt{2}$. Thus $PC = |r| = |a|\sqrt{2}$.For points $A$ and $D$ on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$,substitute $x$ and $y$:$\frac{(a + r/\sqrt{2})^2}{9} + \frac{(2 + r/\sqrt{2})^2}{4} = 1$.Expanding this,we get $4(a^2 + \sqrt{2}ar + r^2/2) + 9(4 + 2\sqrt{2}r + r^2/2) = 36$.$13r^2/2 + (4\sqrt{2}a + 18\sqrt{2})r + 4a^2 = 0$.The product of roots $PA \cdot PD = \frac{4a^2}{13/2} = \frac{8a^2}{13}$.Since $PA, PB, PC, PD$ are in $GP$,$PA \cdot PD = PB \cdot PC$.$\frac{8a^2}{13} = (2\sqrt{2})(|a|\sqrt{2}) = 4|a|$.Assuming $a > 0$,$\frac{8a^2}{13} = 4a \Rightarrow 2a = 13$.

The line through $P(a, 2)$,where $a \neq 0$,making an angle $45^{\circ}$ with the positive direction of the $X$-axis meets the curve $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $A$ and $D$ and the coordinate axes at $B$ and $C$. If $PA, PB, PC$ and $PD$ are in a geometric progression,then $2a=$

Similar Questions

Let $L_1$ be a line passing through the origin and $L_2$ be the line $x + y = 1$. If the intercepts made by the circle $x^{2} + y^{2} - x + 3y = 0$ on $L_1$ and $L_2$ are equal,then which of the following equations represents $L_1$?

Let $a$ and $b$ be non-zero real numbers. Then,the equation $(a x^2+b y^2+c)(x^2-5 x y+6 y^2)=0$ represents

Two chords of the circle $x^2+y^2-2gx-2hy+g^2+h^2-c^2=0$ pass through the point $(g, h+c)$,and the line $y=x$ bisects these two chords. Then:

The tangent at $A(-1, 2)$ on the circle $x^2+y^2-4x-8y+7=0$ touches the circle $x^2+y^2+4x+6y=0$ at $B$. Then,a point of trisection of $AB$ is

Let the parabola $y=x^2+px-3$ meet the coordinate axes at the points $P, Q$ and $R$. If the circle $C$ with center at $(-1,-1)$ passes through the points $P, Q$ and $R$,then the area of $\triangle PQR$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module