If the straight line l x + m y = n is a normal to the curve x = a cos ϕ and y = b sin ϕ , 0 ≤ ϕ ≤ 2 π then prove that a 2 l 2 + b 2 m 2 = a 2 - b 2 2 n 2

Given curve is ellipse. Given that line l x + m y = n is a normal to the curve ellipse at a cos ϕ , b sin ϕ . Now, equation of normal at a cos ϕ , b sin ϕ is a x cos ϕ - b y sin ϕ = a 2 - b 2 Comparing this with given line, we get a l cos ϕ = - b m sin ϕ = a 2 - b 2 n ⇒ cos ϕ = a n l a 2 - b 2 ...(1) ⇒ sin ϕ = - b n m a 2 - b 2 ..(2) Squaring and adding (1), (2), we get a 2 n 2 l 2 a 2 - b 2 2 + b 2 n 2 m 2 a 2 - b 2 2 = 1 ∵ sin 2 ϕ + cos 2 ϕ = 1 ⇒ a 2 l 2 + b 2 m = a 2 - b 2 2 n 2 Hence, proved.

If the line x cos α + y sin α = p touches the ellipse x 2 a 2 + y 2 b 2 = 1 , then prove that a 2 cos 2 α + b 2 sin 2 α = p 2 .

Given, line is x cos α + y sin α = p .......(1) and ellipse is x 2 a 2 + y 2 b 2 = 1 ⇒ b 2 x 2 + a 2 y 2 = a 2 b 2 .......(2) Now, Differentiating the ellipse both the sides w.r.t. x , we get b 2 · 2 x + a 2 · 2 y d y d x = 0 ⇒ d y d x = - x b 2 y a 2 From (1), y sin α = p - x cos α ⇒ y = - x cot α + p sin α Thus slope of line is - cot α . So, the given equation of line (1) will be tangent of the ellipse, if - x b 2 y a 2 = - cot α ⇒ x a 2 cos α = y b 2 sin α = k (say) ⇒ x = k a 2 cos α and y = k b 2 sin α So, the line (1) will touch ellipse at point k a 2 cos α , k b 2 sin α From (1), k a 2 cos 2 α + k b 2 sin 2 α = p [ substitute the point in the line] ⇒ a 2 cos 2 α + b 2 sin 2 α = p k ⇒ a 2 cos 2 α + b 2 sin 2 α 2 = p 2 k 2 ......(3) From (2), b 2 k a 2 cos α 2 + a 2 k b 2 sin α 2 = a 2 b 2 ⇒ k 2 a 2 cos 2 α + b 2 sin 2 α = 1 ⇒ a 2 cos 2 α + b 2 sin 2 α = 1 k 2 ......(4) On dividing Equation (3) by (4), we get a 2 cos 2 α + b 2 sin 2 α = p 2 Hence, if the line touches ellipse, then a 2 cos 2 α + b 2 sin 2 α = p 2 .

If the straight line x cos θ + y sin θ = p touches the curve x a m + y b m = 1 then prove that, a cos θ m m - 1 + b sin θ m m - 1 = p m m - 1 .

The given curve is x a m + y b m = 1 ⇒ m x m - 1 a m + m y m - 1 b m d y d x = 0 [Differentiating both sides w.r.t. x ] ⇒ d y d x = - b a m · x y m - 1 Now, the equation of tangent to curve with slope d y d x at x 1 , y 1 is y - y 1 = - b a m · x 1 y 1 m - 1 x - x 1 ⇒ x 1 m - 1 a m x + y 1 m - 1 b m y = x 1 a m + y 1 b m ⇒ x 1 m - 1 a m x + y 1 m - 1 b m y = 1 [As x 1 , y 1 lies on curve x a m + y b m = 1 ] Given that x cos θ + y sin θ = p is the tangent to curve So, cos θ x 1 m - 1 a m = sin θ y 1 m - 1 b m = p 1 ⇒ a cos θ = p x 1 a m - 1 ⇒ b sin θ = p y 1 b m - 1 Now, ( a cos θ ) m m - 1 + ( b sin θ ) m m - 1 = p m m - 1 · x 1 a m + p m m - 1 · y 1 b m = p m m - 1 x 1 a m + y 1 b m = p m m - 1 1 [As x 1 , y 1 lies on curve] = p m m - 1 Hence, proved that a cos θ m m - 1 + b sin θ m m - 1 = p m m - 1 .

Show that for all nonzero real values of, the straight line y = m ( x + a ) + a m is a tangent to the parabola y 2 = 4 a ( x + a ) .

Given, parabola is y 2 = 4 a ( x + a ) ⇒ 2 y d y d x = 4 a [Differentiating w.r.t. x ] ⇒ d y d x = 2 a y So, slope of tangent to parabola is M = d y d x = 2 a y Given that y = m ( x + a ) + a m is the tangent whose slope is m . So, m = 2 a y [Equating the slopes] ⇒ y = 2 a m ⇒ 2 a m 2 = 4 a x + 4 a 2 [Substituting in parabola] ⇒ 4 a 2 m 2 - 4 a 2 = 4 a x ⇒ x = a m 2 - a Substituting y = 2 a m , x = a m 2 - a in the given line 2 a m = m ( a m 2 - a + a ) + a m ⇒ 2 a m = 2 a m So, the point lies on both curve and given line. Therefore, the given line is tangent to parabola for real all values of m .

Show that for all real values of m , the straight line y = m x + a 2 m 2 - b 2 touches the hyperbola b 2 x 2 - a 2 y 2 = a 2 b 2 .

The given curve is b 2 x 2 - a 2 y 2 = a 2 b 2 ⇒ 2 b 2 x - 2 a 2 y d y d x = 0 [Differentiating both sides w.r.t. x ] ⇒ d y d x = b 2 x a 2 y Now, the equation of tangent to curve with slope d y d x at x 1 , y 1 is y - y 1 = b 2 a 2 x 1 y 1 x - x 1 ⇒ x 1 a 2 x - y 1 b 2 y = x 1 a 2 - y 1 b 2 ⇒ x 1 a 2 x - y 1 b 2 y = 1 [As x 1 , y 1 lies on curve x a 2 - y b 2 = 1 ] Given that y = m x + a 2 m 2 - b 2 is the tangent to curve So, - m x 1 a 2 = 1 - y 1 b 2 = a 2 m 2 - b 2 1 ⇒ x 1 = - m a 2 a 2 m 2 - b 2 ⇒ y 1 = - b 2 a 2 m 2 - b 2 As x 1 , y 1 lies on curve, substituting them in the curve b 2 x 1 2 - a 2 y 1 2 = a 2 b 2 ⇒ b 2 - m a 2 a 2 m 2 - b 2 2 - a 2 - b 2 a 2 m 2 - b 2 2 = a 2 b 2 ⇒ a 2 b 2 m 2 a 2 a 2 m 2 - b 2 - a 2 b 2 b 2 a 2 m 2 - b 2 = a 2 b 2 ⇒ a 2 b 2 m 2 a 2 - b 2 a 2 m 2 - b 2 = a 2 b 2 ⇒ a 2 b 2 = a 2 b 2 So, the point lies on both curve and given line. Therefore, the given line touches the hyperbola for real all values of m .