Let the sum of the focal distances of the point P(4,3) on the hyperbola H: \fracx^2a^2-\fracy^2~b^2=1 be 8 \sqrt\frac53 . If for H , the length of the latus rectum is l and the product of the focal distances of the point P is m , then 9 l^2+6 ~m is equal to :

Hyperbola PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q29 with Solution

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Correct answer: D

Use the focal-distance property of a hyperbola

For the hyperbola

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

let the focal distances of a point $P$ on it be $r_1$ and $r_2$ .

For any point on the hyperbola, the difference of focal distances is:

|r_1-r_2|=2a.

Given also that

r_1+r_2=8\sqrt{\frac53}.

Since $P=(4,3)$ lies on the hyperbola, let us first compute its distance from the center:

OP^2=4^2+3^2=25.

If the foci are $(\pm c,0)$ , then

r_1=\sqrt{(4-c)^2+3^2},\qquad r_2=\sqrt{(4+c)^2+3^2}.

Their product satisfies the identity

(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2.

Now,

r_1^2+r_2^2=((4-c)^2+9)+((4+c)^2+9)=50+2c^2.

Hence

\left(8\sqrt{\frac53}\right)^2=50+2c^2+2r_1r_2.

So,

\frac{320}{3}=50+2c^2+2m

where $m=r_1r_2$ .

But there is a simpler route using sum and difference.

Find the focal distances individually

We know:

r_1+r_2=8\sqrt{\frac53},\qquad r_2-r_1=2a

(for suitable ordering).

Also, using the standard formulas for a point on hyperbola,

r_1r_2=b^2.

Let us derive parameters directly from the point condition.

Since $P(4,3)$ lies on

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

we have

\frac{16}{a^2}-\frac{9}{b^2}=1. \tag{1}

Also for hyperbola,

c^2=a^2+b^2.

Now use the identity for sum of focal distances:

(r_1+r_2)^2=(r_2-r_1)^2+4r_1r_2.

Thus

\left(8\sqrt{\frac53}\right)^2=(2a)^2+4m.

\frac{320}{3}=4a^2+4m

which gives

a^2+m=\frac{80}{3}. \tag{2}

Use the standard focal-distance product formula

For a point $(x,y)$ on the hyperbola

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

the product of distances from foci is

r_1r_2=b^2.

Therefore

m=b^2.

So from (2),

a^2+b^2=\frac{80}{3}.

But

a^2+b^2=c^2.

Hence

c^2=\frac{80}{3}. \tag{3}

Use the actual point coordinates to find $a^2$ and $b^2$

Now compute the sum of squared focal distances another way:

r_1^2+r_2^2=2(OP^2+c^2)=2(25+c^2).

Using (3),

r_1^2+r_2^2=2\left(25+\frac{80}{3}\right)=2\cdot \frac{155}{3}=\frac{310}{3}.

But also,

(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2.

\frac{320}{3}=\frac{310}{3}+2m

which gives

2m=\frac{10}{3}\quad\Rightarrow\quad m=\frac53.

Thus

b^2=\frac53.

Then from (1):

\frac{16}{a^2}-\frac{9}{5/3}=1

\frac{16}{a^2}-\frac{27}{5}=1

\frac{16}{a^2}=\frac{32}{5}

a^2=\frac{5}{2}.

Find latus rectum length

For the hyperbola

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

length of latus rectum is

l=\frac{2b^2}{a}.

l^2=\frac{4b^4}{a^2}.

Now,

b^2=\frac53 \Rightarrow b^4=\frac{25}{9},

therefore

l^2=\frac{4\cdot 25/9}{5/2}=\frac{100}{9}\cdot \frac{2}{5}=\frac{40}{9}.

Hence

9l^2=40.

Also,

6m=6\cdot \frac53=10.

Therefore

9l^2+6m=40+10=50.

Check with options

We obtained

9l^2+6m=50,

but none of the options matches this.

So let us re-check the key formula. For a point on the hyperbola, the product of focal distances is actually

r_1r_2=\left|x^2+y^2-c^2\right|.

Since $P=(4,3)$ and from (3),

m=r_1r_2=25-\frac{80}{3}=-\frac{5}{3}

which is impossible for distances. Thus the previous assumption $r_1r_2=b^2$ was incorrect.

Let us do it correctly.

Correct computation of $m$

For foci $(\pm c,0)$ ,

r_1^2=((4-c)^2+9),\qquad r_2^2=((4+c)^2+9).

Thus

r_1^2r_2^2=\big((4-c)^2+9\big)\big((4+c)^2+9\big).

This simplifies as

((25+c^2)^2-(8c)^2)=(25-c^2)^2+24c^2.

Now from the identity

(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2,

we get

4a^2=\frac{320}{3}-4m.

a^2+m=\frac{80}{3}. \tag{4}

Also since $m=r_1r_2$ , and using

r_1^2+r_2^2=50+2c^2,

we have

\frac{320}{3}=50+2c^2+2m

\Rightarrow c^2+m=\frac{85}{3}. \tag{5}

But

c^2=a^2+b^2.

Subtracting (4) from (5):

b^2=\frac{5}{3}.

Then from the point condition,

\frac{16}{a^2}-\frac{9}{5/3}=1 Rightarrow \frac{16}{a^2}-\frac{27}{5}=1 Rightarrow \frac{16}{a^2}=\frac{32}{5} Rightarrow a^2=\frac{5}{2}.

Hence

c^2=a^2+b^2=\frac{5}{2}+\frac{5}{3}=\frac{25}{6}.

Now from (5),

m=\frac{85}{3}-\frac{25}{6}=\frac{170-25}{6}=\frac{145}{6}.

Final calculation

Again,

l=\frac{2b^2}{a},

l^2=\frac{4b^4}{a^2}=\frac{4\cdot (25/9)}{5/2}=\frac{40}{9}.

Thus

9l^2=40.

And

6m=6\cdot \frac{145}{6}=145.

Therefore

9l^2+6m=40+145=185.

So the correct option is:

\boxed{185}

which is Option D.

More from Hyperbola

View written solutionFree

Correct answer: D

Use the focal-distance property of a hyperbola

For the hyperbola

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

let the focal distances of a point $P$ on it be $r_1$ and $r_2$ .

For any point on the hyperbola, the difference of focal distances is:

|r_1-r_2|=2a.

Given also that

r_1+r_2=8\sqrt{\frac53}.

Since $P=(4,3)$ lies on the hyperbola, let us first compute its distance from the center:

OP^2=4^2+3^2=25.

If the foci are $(\pm c,0)$ , then

r_1=\sqrt{(4-c)^2+3^2},\qquad r_2=\sqrt{(4+c)^2+3^2}.

Their product satisfies the identity

(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2.

Now,

r_1^2+r_2^2=((4-c)^2+9)+((4+c)^2+9)=50+2c^2.

Hence

\left(8\sqrt{\frac53}\right)^2=50+2c^2+2r_1r_2.

So,

\frac{320}{3}=50+2c^2+2m

where $m=r_1r_2$ .

But there is a simpler route using sum and difference.

Find the focal distances individually

We know:

r_1+r_2=8\sqrt{\frac53},\qquad r_2-r_1=2a

(for suitable ordering).

Also, using the standard formulas for a point on hyperbola,

r_1r_2=b^2.

Let us derive parameters directly from the point condition.

Since $P(4,3)$ lies on

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

we have

\frac{16}{a^2}-\frac{9}{b^2}=1. \tag{1}

Also for hyperbola,

c^2=a^2+b^2.

Now use the identity for sum of focal distances:

(r_1+r_2)^2=(r_2-r_1)^2+4r_1r_2.

Thus

\left(8\sqrt{\frac53}\right)^2=(2a)^2+4m.

\frac{320}{3}=4a^2+4m

which gives

a^2+m=\frac{80}{3}. \tag{2}

Use the standard focal-distance product formula

For a point $(x,y)$ on the hyperbola

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

the product of distances from foci is

r_1r_2=b^2.

Therefore

m=b^2.

So from (2),

a^2+b^2=\frac{80}{3}.

But

a^2+b^2=c^2.

Hence

c^2=\frac{80}{3}. \tag{3}

Use the actual point coordinates to find $a^2$ and $b^2$

Now compute the sum of squared focal distances another way:

r_1^2+r_2^2=2(OP^2+c^2)=2(25+c^2).

Using (3),

r_1^2+r_2^2=2\left(25+\frac{80}{3}\right)=2\cdot \frac{155}{3}=\frac{310}{3}.

But also,

(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2.

\frac{320}{3}=\frac{310}{3}+2m

which gives

2m=\frac{10}{3}\quad\Rightarrow\quad m=\frac53.

Thus

b^2=\frac53.

Then from (1):

\frac{16}{a^2}-\frac{9}{5/3}=1

\frac{16}{a^2}-\frac{27}{5}=1

\frac{16}{a^2}=\frac{32}{5}

a^2=\frac{5}{2}.

Find latus rectum length

For the hyperbola

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

length of latus rectum is

l=\frac{2b^2}{a}.

l^2=\frac{4b^4}{a^2}.

Now,

b^2=\frac53 \Rightarrow b^4=\frac{25}{9},

therefore

l^2=\frac{4\cdot 25/9}{5/2}=\frac{100}{9}\cdot \frac{2}{5}=\frac{40}{9}.

Hence

9l^2=40.

Also,

6m=6\cdot \frac53=10.

Therefore

9l^2+6m=40+10=50.

Check with options

We obtained

9l^2+6m=50,

but none of the options matches this.

So let us re-check the key formula. For a point on the hyperbola, the product of focal distances is actually

r_1r_2=\left|x^2+y^2-c^2\right|.

Since $P=(4,3)$ and from (3),

m=r_1r_2=25-\frac{80}{3}=-\frac{5}{3}

which is impossible for distances. Thus the previous assumption $r_1r_2=b^2$ was incorrect.

Let us do it correctly.

Correct computation of $m$

For foci $(\pm c,0)$ ,

r_1^2=((4-c)^2+9),\qquad r_2^2=((4+c)^2+9).

Thus

r_1^2r_2^2=\big((4-c)^2+9\big)\big((4+c)^2+9\big).

This simplifies as

((25+c^2)^2-(8c)^2)=(25-c^2)^2+24c^2.

Now from the identity

(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2,

we get

4a^2=\frac{320}{3}-4m.

a^2+m=\frac{80}{3}. \tag{4}

Also since $m=r_1r_2$ , and using

r_1^2+r_2^2=50+2c^2,

we have

\frac{320}{3}=50+2c^2+2m

\Rightarrow c^2+m=\frac{85}{3}. \tag{5}

But

c^2=a^2+b^2.

Subtracting (4) from (5):

b^2=\frac{5}{3}.

Then from the point condition,

\frac{16}{a^2}-\frac{9}{5/3}=1 Rightarrow \frac{16}{a^2}-\frac{27}{5}=1 Rightarrow \frac{16}{a^2}=\frac{32}{5} Rightarrow a^2=\frac{5}{2}.

Hence

c^2=a^2+b^2=\frac{5}{2}+\frac{5}{3}=\frac{25}{6}.

Now from (5),

m=\frac{85}{3}-\frac{25}{6}=\frac{170-25}{6}=\frac{145}{6}.

Final calculation

Again,

l=\frac{2b^2}{a},

l^2=\frac{4b^4}{a^2}=\frac{4\cdot (25/9)}{5/2}=\frac{40}{9}.

Thus

9l^2=40.

And

6m=6\cdot \frac{145}{6}=145.

Therefore

9l^2+6m=40+145=185.

So the correct option is:

\boxed{185}

which is Option D.

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