A line segment joining a point A on x-axis to | vedclass

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

Question

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Given $AB = 15$,so $\sqrt{a^2 + b^2} = 15$,which implies $a^2 + b^2 = 225$.Let $P(x, y)

Vedclass · Accepted Answer

$x=9 \cos 	heta, y=6 \sin 	heta$

Vedclass · Answer

(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Given $AB = 15$,so $\sqrt{a^2 + b^2} = 15$,which implies $a^2 + b^2 = 225$.Let $P(x, y)$ be a point on $AB$ such that $\frac{AP}{PB} = \frac{2}{3}$.Using the section formula,$x = \frac{2(0) + 3(a)}{2+3} = \frac{3a}{5}$ and $y = \frac{2(b) + 3(0)}{2+3} = \frac{2b}{5}$.From these,$a = \frac{5x}{3}$ and $b = \frac{5y}{2}$.Substituting these into $a^2 + b^2 = 225$:$(\frac{5x}{3})^2 + (\frac{5y}{2})^2 = 225$.$\frac{25x^2}{9} + \frac{25y^2}{4} = 225$.Dividing by $25$,we get $\frac{x^2}{9} + \frac{y^2}{4} = 9$,or $\frac{x^2}{81} + \frac{y^2}{36} = 1$.This represents an ellipse with $x = 9 \cos 	heta$ and $y = 6 \sin 	heta$.

$A$ line segment joining a point $A$ on $x$-axis to a point $B$ on $y$-axis is such that $AB=15$. If $P$ is a point on $AB$ such that $\frac{AP}{PB}=\frac{2}{3}$,then the locus of $P$ is:

Similar Questions

$A$ straight line $L$ cuts both the lines $5x - y - 4 = 0$ and $3x + 4y - 4 = 0$. The segment of $L$ between the two lines is bisected at the point $(1, 5)$. The equation of $L$ is

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA$ is equal to:

If $A(1,0), B(0,-2), C(2,-1)$ are three fixed points,then the equation of the locus of a point $P(x,y)$ such that the area of $\triangle PAB$ is equal to the area of $\triangle PAC$ is:

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

The locus of a point which moves so that it is always equidistant from the points $A(a, 0)$ and $B(-a, 0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module