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Byju's Answer

Standard XII

Mathematics

Definition of Ellipse

A straight li...

Question

A straight line through A(6, 8) meets the curve $2 x^{2} + y^{2} = 2$ at B and C. P is a point on BC such that AB, AP, AC are in H.P, then the minimum distance of the origin from the locus of ‘P’ is

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Solution

The correct option is A
$(6 + r c o s θ, 8 + r s i n θ)$ lies on $2 x^{2} + y^{2} = 2$
$\Rightarrow (2 c o s^{2} θ + s i n^{2} θ) r^{2} + 2 (12 c o s θ + 8 s i n θ) r + 134 = 0$
AB, AP, AC are in H.P $\Rightarrow \frac{2}{r} = \frac{A B + A C}{A B . A C} \Rightarrow \frac{1}{r} = - \frac{6 c o s θ + 4 s i n θ}{67} \Rightarrow 6 x + 4 y - 1 = 0$
Minimum distance from 'O'= $\frac{1}{\sqrt{52}}$

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A straight line through A(6, 8) meets the curve 2x2+y2=2 at B and C. P is a point on BC such that AB, AP, AC are in H.P, then the minimum distance of the origin from the locus of ‘P’ is

The correct option is A (6+rcos θ,8+rsin θ) lies on 2x2+y2=2 ⇒(2 cos2 θ+sin2 θ)r2+2(12cos θ+8sin θ) r+134=0 AB, AP, AC are in H.P ⇒2r=AB+ACAB.AC⇒1r=−6cos θ+4sin θ67⇒6x+4y−1=0 Minimum distance from 'O'= 1√52

A straight line through A(6, 8) meets the curve $2 x^{2} + y^{2} = 2$ at B and C. P is a point on BC such that AB, AP, AC are in H.P, then the minimum distance of the origin from the locus of ‘P’ is