Ellipse (S.N.Dey) |Part-8 |Ex-5

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In the previous article , we solved few solutions of Long Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11. In this chapter, we will solve few more.
S N Dey-Ellipse-solutions
Ellipse

8. ~PQ~ is any double ordinate of the ellipse ~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~; find the equation to the locus of the point of trisection of ~PQ~ that is nearer to ~P.

Solution.

Let ~P\equiv (a\cos\theta, b\sin\theta),~Q\equiv(a\cos\theta,-b\sin\theta).

Suppose that the point ~R(h,k)~ is the nearest point of P which is the point of trisection of ~PQ~ i.e., ~PR : RQ=1:2.

\therefore~ h=\frac{1 \cdot a\cos\theta+2 \cdot a\cos\theta}{1+2}=\frac{3a\cos\theta}{3} =a\cos\theta \\ \text{or,}~~ \cos\theta=\frac ha\rightarrow(1)\\~ k=\frac{(-b\sin\theta)\times 1+2 \times b\sin\theta}{1+2}=\frac{b\sin\theta}{3} \\ \text{or,}~~ \sin\theta=\frac{3k}{b}\rightarrow(2)

\because~ \sin^2\theta+\cos^2\theta=1 \\ \text{or,}~~ \left(\frac{3k}{b}\right)^2+\left(\frac ha\right)^2=1~~[\text{By (1),(2)}] \\ \text{or,}~~ \frac{9k^2}{b^2}+\frac{h^2}{a^2}=1 \\ \text{or,}~~ \frac{h^2}{a^2}+\frac{9k^2}{b^2}=1\rightarrow(3)

Hence, from (3) we get the locus of the point which is ~\frac{x^2}{a^2}+\frac{9y^2}{b^2}=1.

For Full Solution PDF of the Ellipse (S N De-Chhaya Mathematics), click here.

9. Show that the double ordinate of the auxiliary circle  of an ellipse passing through the focus is equal to the minor axis of the ellipse.

Solution.

Let the equation of the ellipse be 

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ~~(a^2>b^2)\rightarrow(1)

and the auxiliary circle of (1) is ~x^2+y^2=a^2\rightarrow(2)

Suppose that PP' is the double ordinate of the auxiliary circle (2).

\therefore~ P \equiv(a\cos\theta,a\sin\theta),~~P'\equiv (a\cos\theta,-a\sin\theta), Q \equiv (a\cos\theta,b\sin\theta),\\~~Q'\equiv (a\cos\theta,-b\sin\theta).

Now, the equation of QQ' is given by ~x-a\cos\theta=0.

Since QQ' passes through the focus i.e., (ae,0), so

~ae-a\cos\theta=0 \Rightarrow a(e-\cos\theta)=0 \Rightarrow \cos\theta=e.

Since the auxiliary circle has double ordinate, 

~PP'=2a\sin\theta=2a\sqrt{1-\cos^2\theta}=2a\sqrt{1-e^2}=2a\cdot \sqrt{\frac{b^2}{a^2}}(*)\\ \therefore~ PP'=2a \times \frac ba=2b=\text{length of minor axis}.

Note[*] : ~ e^2=1-\frac{b^2}{a^2} \Rightarrow \frac{b^2}{a^2}=1-e^2.

10. ~O~ is the centre of an ellipse whose semi-minor axis is ~b. The ordinate of a point ~P~ of the ellipse intersects its auxiliary circle at ~Q~ (when produced). The straight line through ~P~  drawn parallel to ~OQ~ cuts the major axis at ~G.~ Prove that, ~PG=b.

Solution.

Let the equation of the ellipse be ~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.

Clearly,~~\Delta GNP~ and ~\Delta ONQ~ are similar triangle,

\therefore~ \frac{PG}{OQ}=\frac{PN}{QN} \\ \text{or,}~~ \frac{PG}{\sqrt{a^2\cos^2\theta+a^2\sin^2\theta}}=\frac{b\sin\theta}{a\sin\theta} \\ \text{or,}~~ \frac{PG}{a}=\frac ba \\ \therefore~ PG=b.

11. If ~(\alpha+\beta)~ and ~(\alpha-\beta)~ are the eccentric  angles of the points ~P~ and ~Q~ respectively on the ellipse ~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~, show that the equation of the chord ~PQ~ is 

\frac xa \cos\alpha+\frac yb\sin\alpha=\cos\beta.

Solution.

Since ~(\alpha+\beta)~,~(\alpha-\beta)~ are the eccentric angles of the points ~P~ and ~Q~respectively, so

~P\equiv (a\cos(\alpha+\beta),b\sin(\alpha+\beta)),~ Q\equiv (a\cos(\alpha-\beta),b\sin(\alpha-\beta)).

\therefore~ The equation of PQ is 

~\frac{y-a\sin(\alpha+\beta)}{x-a\cos(\alpha+\beta)}=\frac{b\sin(\alpha-\beta)-b\sin(\alpha+\beta)}{a\cos(\alpha-\beta)-a\cos(\alpha+\beta)} \\ \text{or,}~~ \frac{y-b\sin(\alpha+\beta)}{x-a\cos(\alpha+\beta)}=\frac ba \cdot \frac{-2\cos\alpha \sin\beta}{2\sin\alpha \sin\beta} \\ \text{or,}~~ \frac{y-b\sin(\alpha+\beta)}{x-a\cos(\alpha+\beta)}=-\frac ba \cdot \frac{\cos\alpha}{\sin\alpha} \\ \text{or,}~~ bx \cos\alpha+ay\sin\alpha=ab[\cos(\alpha+\beta)\cos\alpha+\sin(\alpha+\beta)\sin\alpha] \\ \text{or,}~~ bx\cos\alpha+ay\sin\alpha=ab\cos\beta \\ \therefore~ \frac xa \cos\alpha+\frac yb \sin\alpha=\cos\beta. 

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