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Question: Let H=(x-1)^2/9-(y+2)^2/4=1 be hyperbola wit cnetre at C. from point Q on the asymptotes a line is d...

Let H=(x-1)^2/9-(y+2)^2/4=1 be hyperbola wit cnetre at C. from point Q on the asymptotes a line is drawn perpendiular to transverse axis that meet sparabola at P and P'. Again, a tangent is drawn to hyperbola at P which meets asymptotes at L and G. find PQ.P'Q and area of triangle CLG

Answer

PQ * P'Q = 4, Area of triangle CLG = 6

Explanation

Solution

Solution

We are given the hyperbola

(x−1)29−(y+2)24=1,\frac{(x-1)^2}{9} - \frac{(y+2)^2}{4} = 1,

with centre C=(1,−2)C=(1,-2) and transverse axis horizontal. Its asymptotes are

y+2=±23(x−1).y+2=\pm\frac{2}{3}(x-1).

Let QQ be a point on one of the asymptotes, say on

y+2=23(x−1).y+2=\frac{2}{3}(x-1).

So we may take

Q=(q,  23(q−1)−2).Q=(q,\; \frac{2}{3}(q-1)-2).

A line through QQ perpendicular to the transverse axis (i.e. vertical, x=qx=q) meets the hyperbola. Substituting x=qx=q in the hyperbola we have

(q−1)29−(y+2)24=1.\frac{(q-1)^2}{9} -\frac{(y+2)^2}{4} =1.

Rearrange to get

(y+2)24=(q−1)2−99⟹(y+2)2=4((q−1)2−9)9.\frac{(y+2)^2}{4}=\frac{(q-1)^2-9}{9}\quad \Longrightarrow \quad (y+2)^2=\frac{4((q-1)^2-9)}{9}.

Thus the two points of intersection (call them PP and P′P') are

y+2=±23(q−1)2−9.y+2=\pm\frac{2}{3}\sqrt{(q-1)^2-9}.

That is,

P=(q, −2+23(q−1)2−9),P′=(q, −2−23(q−1)2−9).P = \Bigl(q,\,-2+\frac{2}{3}\sqrt{(q-1)^2-9}\Bigr),\quad P' = \Bigl(q,\,-2-\frac{2}{3}\sqrt{(q-1)^2-9}\Bigr).

Since the vertical line makes the distances PQPQ and P′QP'Q simply the vertical differences, we find

PQ=∣[−2+23(q−1)2−9]−[23(q−1)−2]∣=23∣(q−1)2−9−(q−1)∣,P′Q=∣[−2−23(q−1)2−9]−[23(q−1)−2]∣=23∣(q−1)2−9+(q−1)∣.\begin{array}{rcl} PQ &=& \left|\left[-2+\frac{2}{3}\sqrt{(q-1)^2-9}\right]-\left[\frac{2}{3}(q-1)-2\right]\right| =\frac{2}{3}\Bigl|\sqrt{(q-1)^2-9}-(q-1)\Bigr|,\\[1mm] P'Q &=& \left|\left[-2-\frac{2}{3}\sqrt{(q-1)^2-9}\right]-\left[\frac{2}{3}(q-1)-2\right]\right| =\frac{2}{3}\Bigl|\sqrt{(q-1)^2-9}+(q-1)\Bigr|. \end{array}

Thus, their product is

PQ⋅P′Q=(23)2∣(q−1)2−9−(q−1)∣∣(q−1)2−9+(q−1)∣.PQ\cdot P'Q=\left(\frac{2}{3}\right)^2\Bigl|\sqrt{(q-1)^2-9}-(q-1)\Bigr|\Bigl|\sqrt{(q-1)^2-9}+(q-1)\Bigr|.

But note that

∣(q−1)2−9−(q−1)∣∣(q−1)2−9+(q−1)∣=∣((q−1)2−9)2−(q−1)2∣=∣(q−1)2−9−(q−1)2∣=9.\Bigl|\sqrt{(q-1)^2-9}-(q-1)\Bigr|\Bigl|\sqrt{(q-1)^2-9}+(q-1)\Bigr|=\Bigl|(\sqrt{(q-1)^2-9})^2-(q-1)^2\Bigr|=|(q-1)^2-9-(q-1)^2|=9.

Hence,

PQ⋅P′Q=49⋅9=4.PQ\cdot P'Q=\frac{4}{9}\cdot9=4.

Next, let’s find the area of △CLG\triangle CLG where LL and GG are the intersections of the tangent to the hyperbola at PP with the asymptotes.

Step 1. Tangent at PP:

For the hyperbola

(x−1)29−(y+2)24=1,\frac{(x-1)^2}{9} - \frac{(y+2)^2}{4} = 1,

the tangent at a point (x0,y0)(x_0,y_0) on it is

(x0−1)(x−1)9−(y0+2)(y+2)4=1.\frac{(x_0-1)(x-1)}{9} - \frac{(y_0+2)(y+2)}{4}=1.

At P=(q, −2+23(q−1)2−9)P=(q,\, -2+\frac{2}{3}\sqrt{(q-1)^2-9}) we note that

y0+2=23(q−1)2−9.y_0+2=\frac{2}{3}\sqrt{(q-1)^2-9}.

Let D=q−1D=q-1 and S=D2−9S=\sqrt{D^2-9}. Then the tangent becomes

D(x−1)9−(23S)(y+2)4=1.\frac{D (x-1)}{9} - \frac{\left(\frac{2}{3}S\right)(y+2)}{4}=1.

Simplify the coefficient of (y+2)(y+2):

23⋅14=16.\frac{2}{3}\cdot\frac{1}{4} = \frac{1}{6}.

Thus the tangent equation is

D9(x−1)−S6(y+2)=1.\frac{D}{9}(x-1) - \frac{S}{6}(y+2)=1.

Step 2. Find LL and GG:

The asymptotes are:

  • y+2=23(x−1)y+2=\frac{2}{3}(x-1)   (say intersection point LL)
  • y+2=−23(x−1)y+2=-\frac{2}{3}(x-1)   (say intersection point GG)

For LL:

Substitute y+2=23(x−1)y+2=\frac{2}{3}(x-1) in the tangent:

D9(x−1)−S6⋅23(x−1)=1.\frac{D}{9}(x-1)-\frac{S}{6}\cdot\frac{2}{3}(x-1)=1.

Simplify:

(x−1)(D9−2S18)=(x−1)(D−S9)=1.(x-1)\left(\frac{D}{9}-\frac{2S}{18}\right)= (x-1)\left(\frac{D-S}{9}\right)=1.

Thus,

x−1=9D−S.x-1=\frac{9}{D-S}.

Then,

y+2=23⋅9D−S=6D−S.y+2=\frac{2}{3}\cdot\frac{9}{D-S}=\frac{6}{D-S}.

So,

L=(1+9D−S,  −2+6D−S).L=\left(1+\frac{9}{D-S},\;-2+\frac{6}{D-S}\right).

For GG:

Substitute y+2=−23(x−1)y+2=-\frac{2}{3}(x-1) in the tangent:

D9(x−1)−S6[−23(x−1)]=D9(x−1)+S9(x−1)=(x−1)D+S9=1.\frac{D}{9}(x-1)-\frac{S}{6}\Bigl[-\frac{2}{3}(x-1)\Bigr] = \frac{D}{9}(x-1)+\frac{S}{9}(x-1)= (x-1)\frac{D+S}{9}=1.

Thus,

x−1=9D+S.x-1=\frac{9}{D+S}.

And,

y+2=−23⋅9D+S=−6D+S.y+2=-\frac{2}{3}\cdot\frac{9}{D+S}=-\frac{6}{D+S}.

So,

G=(1+9D+S,  −2−6D+S).G=\left(1+\frac{9}{D+S},\;-2-\frac{6}{D+S}\right).

Step 3. Area of △CLG\triangle CLG:

The vertices of the triangle are:

C=(1,−2),L=(1+9D−S, −2+6D−S),G=(1+9D+S, −2−6D+S).C=(1,-2),\quad L=\left(1+\frac{9}{D-S},\,-2+\frac{6}{D-S}\right),\quad G=\left(1+\frac{9}{D+S},\,-2-\frac{6}{D+S}\right).

Form the vectors from CC to LL and GG:

CL→=(9D−S,  6D−S),CG→=(9D+S,  −6D+S).\overrightarrow{CL}=\left(\frac{9}{D-S},\;\frac{6}{D-S}\right),\quad \overrightarrow{CG}=\left(\frac{9}{D+S},\;-\frac{6}{D+S}\right).

The area is

Area=12∣det⁡(CL→,CG→)∣.\text{Area}=\frac{1}{2}\Bigl|\det(\overrightarrow{CL},\overrightarrow{CG})\Bigr|.

Compute the determinant:

det⁡=9D−S⋅(−6D+S)−6D−S⋅9D+S=−54(D−S)(D+S)−54(D−S)(D+S)=−108D2−S2.\det=\frac{9}{D-S}\cdot\left(-\frac{6}{D+S}\right)-\frac{6}{D-S}\cdot\frac{9}{D+S} = -\frac{54}{(D-S)(D+S)}-\frac{54}{(D-S)(D+S)}=-\frac{108}{D^2-S^2}.

But

D2−S2=(q−1)2−((q−1)2−9)=9.D^2-S^2=(q-1)^2-( (q-1)^2-9)=9.

Thus,

det⁡=−1089=−12.\det=-\frac{108}{9}=-12.

So the area is

Area=12×12=6.\text{Area}=\frac{1}{2}\times 12 = 6.

Final Answers:

  • PQ⋅P′Q=4PQ\cdot P'Q=4
  • Area△CLG=6\triangle CLG=6 square units