If the portion of the line lx + my = 1 fallin | vedclass

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

Question

(C) To find the angle subtended by the chord at the origin,we homogenize the equation of the circle ${x^2} + {y^2} = {a^2}$ using the line $lx + my =

Vedclass · Accepted Answer

$4[{a^2}({l^2} + {m^2}) - 1] = {[{a^2}({l^2} + {m^2}) - 2]^2}$

Vedclass · Answer

(C) To find the angle subtended by the chord at the origin,we homogenize the equation of the circle ${x^2} + {y^2} = {a^2}$ using the line $lx + my = 1$.${x^2} + {y^2} = {a^2}{(lx + my)^2}$${x^2} + {y^2} = {a^2}({l^2}{x^2} + {m^2}{y^2} + 2lmxy)$$({a^2}{l^2} - 1){x^2} + 2{a^2}lmxy + ({a^2}{m^2} - 1){y^2} = 0$This is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = {a^2}{l^2} - 1$,$H = {a^2}lm$,and $B = {a^2}{m^2} - 1$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{H^2 - AB}}{A + B} ight|$.Given $	heta = 45^\circ$,so $	an 45^\circ = 1$.$1 = \left| \frac{2\sqrt{({a^2}lm)^2 - ({a^2}{l^2} - 1)({a^2}{m^2} - 1)}}{{a^2}{l^2} - 1 + {a^2}{m^2} - 1} ight|$$|{a^2}({l^2} + {m^2}) - 2| = 2\sqrt{{a^4}{l^2}{m^2} - ({a^4}{l^2}{m^2} - {a^2}{l^2} - {a^2}{m^2} + 1)}$$|{a^2}({l^2} + {m^2}) - 2| = 2\sqrt{{a^2}({l^2} + {m^2}) - 1}$Squaring both sides,we get:$[{a^2}({l^2} + {m^2}) - 2]^2 = 4[{a^2}({l^2} + {m^2}) - 1]$

If the portion of the line $lx + my = 1$ falling inside the circle ${x^2} + {y^2} = {a^2}$ subtends an angle of $45^\circ$ at the origin,then

Similar Questions

The equation of a common tangent to the circle $x^2+y^2=16$ and the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ is

The values of the constant term in the equation of a circle passing through $(1, 2)$ and $(3, 4)$ and touching the line $3x + y - 3 = 0$ are

The vertex of the parabola $(y - 1)^2 = 8(x - 1)$ is at the centre of a circle and the parabola cuts that circle at the ends of its latus rectum. Then the equation of that circle is

Let $y^2 = 4ax$ be a parabola and $x^2 + y^2 + 2bx = 0$ be a circle. If the parabola and the circle touch each other externally,then:

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 $2a=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module