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Byju's Answer

Standard XII

Mathematics

Solution of Triangle

Let Ω be an...

Question

Let $Ω$ be an ellipse and let $F_{1}$ and $F_{2}$ be its two foci. Let $P$ be a point on the circumference of the ellipse that is neither on the major axis nor on the minor axis. Let $P F_{1}$ intersect $Ω$ again at $Q (\neq P)$ and let $P F_{2}$ intersect $Ω$ again at $R (\neq P$ ). Then

The perimeters of the triangles

P F_{1} R

and

P F_{2} Q

are equal

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The area of the triangle

P F_{1} R

and

P F_{2} Q

are equal if and only if

F_{1} F_{2}

is parallel to

Q R

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The perimeters of the triangles

P F_{1} R

and

P F_{2} Q

are bisected by

F_{1} F_{2}

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The two lines

F_{1} F_{2}

and

Q R

are always parallel

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Solution

The correct options are
A The perimeters of the triangles $P F_{1} R$ and $P F_{2} Q$ are equal
B The area of the triangle $P F_{1} R$ and $P F_{2} Q$ are equal if and only if $F_{1} F_{2}$ is parallel to $Q R$
C The perimeters of the triangles $P F_{1} R$ and $P F_{2} Q$ are bisected by $F_{1} F_{2}$

Let $Ω$ be a standard ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

IN the figure the standard ellipse $Ω$ is drawn with two foci $F_{1}$ and $F_{2}$ . $P$ is a point on the circumference when extended through $F_{1}$ meets the ellipse again at $Q$ and when extended through $F_{2}$ meets the ellipse again at $R$ .

We know that for any ellipse the sum of distances of any point from both foci is always equal to length of major axis i,e. $2 a$ .

Hence for points $P, Q$ and $R$ ,

$P F_{1} + P F_{2} = 2 a$ ... $(1)$

$Q F_{1} + Q F_{2} = 2 a$ ... $(2)$

$R F_{1} + R F_{2} = 2 a$ ... $(3)$
Now if take triangles $P F_{1} R$ and $P F_{2} Q$ , then the perimeters of these triangles are:

For $P F_{1} R$ $\to S_{1} = P F_{1} + R F_{1} + P R$

For $P F_{2} Q$ $\to S_{2} = P F_{2} + Q F_{2} + P Q$

$P R = P F_{2} + R F_{2}$

$P Q = P F_{1} + Q F_{1}$

Now putting values of $P R$ and $P Q$ in $S_{1}$ and $S_{2}$ , we get,

$S_{1} = P F_{1} + P F_{2} + R F_{1} + R F_{2} = (P F_{1} + P F_{2}) + (R F_{1} + R F_{2})$

$S_{2} = P F_{2} + P F_{1} + Q F_{1} + Q F_{2} = (P F_{1} + P F_{2}) + (Q F_{1} + Q F_{2})$

Now putting values from Eq $(1), (2)$ and $(3)$ ,

$S_{1} = 4 a$

$S_{2} = 4 a$

Hence we can see that $S_{1} = S_{2}$ , So the perimeters of trainagles $P F_{1} R$ and $P F_{2} Q$ are equal.

From Eq. $(1), (2)$ and $(3)$ , we can see that,

$P F_{1} + P F_{2} = 2 a$ ... $(1)$

$Q F_{1} + Q F_{2} = 2 a$ ... $(2)$

$R F_{1} + R F_{2} = 2 a$ ... $(3)$

$F_{1} F_{2}$ cuts the perimeter of triangle $P F_{2} Q$ into two parts.

As $P F_{1} + P F_{2} = 2 a$ $= Q F_{1} + Q F_{2}$

So we can say $F_{1} F_{2}$ bisects the perimeter of triangle $P F_{2} Q$

Also $F_{1} F_{2}$ cuts the perimeter of triangle $P F_{1} R$ into two parts too.

As $P F_{1} + P F_{2} = 2 a$ $= R F_{1} + R F_{2}$

So we can also say $F_{1} F_{2}$ bisects the perimeter of triangle $P F_{1} R$

Hence the correct options are $A$ , $B$ and $C$ .

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Similar questions

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Let Ω be an ellipse and let F1 and F2 be its two foci. Let P be a point on the circumference of the ellipse that is neither on the major axis nor on the minor axis. Let PF1 intersect Ω again at Q(≠P) and let PF2 intersect Ω again at R(≠P). Then

Let $Ω$ be an ellipse and let $F_{1}$ and $F_{2}$ be its two foci. Let $P$ be a point on the circumference of the ellipse that is neither on the major axis nor on the minor axis. Let $P F_{1}$ intersect $Ω$ again at $Q (\neq P)$ and let $P F_{2}$ intersect $Ω$ again at $R (\neq P$ ). Then