Tangents and Normals to Conics MCQ Quiz in తెలుగు - Objective Question with Answer for Tangents and Normals to Conics - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

y = x³ - 6x² + 9x + 4

To get an equation of tangent, differentiative the curve and equal it zero.

y' = 3x² - 12x + 9

Now, y' = 0

x = 1, 3

Checking the value of y' at boundary points,

f(1) = 1 - 6 + 9 + 4

= 14 - 6 = 8

f(3) = 27 - 54 + 27 + 4 = 4

f(5) = 125 - 150 + 45 + 4

= 24

Thus the maximum value of slope will be at x = 5

Concept: 

Two curves intersect at right angles means they are orthogonal :

Product of Slope of both curves = -1

Calculation

\( \frac{x^2}{9} + \frac{y^2}{α } = 1 \)

Differentiating with respect to x:

⇒ Slope = y' = \( -\frac{x}{9} × \frac{α}{y} \)

\(y^3 = 81x \)

Differentiating with respect to x:

⇒ ​Slope = \( y' = \frac{27}{y^2}\) 

Product of Slope of both curves

⇒ ​\( -\frac{x}{9} × \frac{α}{y} \) × \(\frac{27}{y^2} \) = -1

⇒ ​α = 27

Hence Option(1) is the correct answer.

Explanation -

3x+4y =12√2 is a tangent to the ellipse (x2/a2) + (y2/9) = 1

Equation of tangent to ellipse (x2/a2) + (y2/9) = 1 is y = mx + √(a2 m2+ 9)

Now, 3x + 4y = 12√2 ⇒ y = -(3/4)x + 3√2

m = -3/4

And √(a2 m2+ 9) = 3√2

(a2 (-3/4)2+ 9) = 18

a2 (9/16) = 9

a2 = 16

a = 4

e = √(1-b2/a2)

e = √(1-9/16)

= √7/4

Distance between foci is 2ae = 2 × 4 × √7/4

= 2√7

Hence the Correct option is (2)

Calculation:

Equation of hyperbola is \(\frac{x^2}{16} - \frac{y^2}{9} = 1\)

Let (h, k) be the midpoint of the chord of the circle x² + y² = 16 .

∴ The equation of the chord will be hx + ky = h² + k²  

⇒ \(y = \frac{-h}{k}x + \frac{h^2 + k^2}{k}\)

This will touch the circle if c² = a²m² - b² 

⇒ \(\left( \frac{h^2 + k^2}{k} \right) = 16 \left( \frac{-h}{k} \right)^2 - 9\)

⇒ (h² + k²)² = 16h² − 9k²

Replacing h by x and y by k we get, (x2 + y2)2 = 16x2 – 9y2​

∴ Required locus is (x2 + y2)2 = 16x2 – 9y2​.

The correct answer is Option 1.

Calculation:

Given, y2 = 4ax 

The coordinates of the focus of the parabola y² = 4ax are (a, 0).

Now, the line x – y – a = 0 passes through this point.

⇒ It is a focal chord of the parabola.

⇒ Tangent intersects at right angle.

∴ The angle between the tangents to the parabola y2 = 4ax at the points where it intersects with the line x – y – a = 0 is \(\frac{\pi}{2}\).

The correct answer is Option 4.

Concept:

The equation of tangent on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) is given by y = mx + \(\sqrt{a^2m^2+b^2}\)

Calculation:

GIven, x2 + 3y2 = 6 is an ellipse.

⇒ \(\frac{x^2}{6}+\frac{y^2}{3}=1\)

∴ Equation of tangent is y = mx + \(\sqrt{6m^2+2}\) … (i)

∴ Equation of line passing through (0, 0) and perpendicular to (i) is:

\(\frac{y-0}{x-0}=-\frac{1}{m}\)

⇒ m = \(-\frac{x}{y}\)

∴ Putting the value of m in (i), we get:

y = \((-\frac{x}{y})\)x + \(\sqrt{6\left(\frac{-x}{y}\right)^2+2}\)

⇒ y² = - x² + \(\sqrt{6x^2+2y^2}\)

⇒ x² + y² = \(\sqrt{6x^2+2y^2}\)

⇒ (x2 + y2)2 = 6x2 + 2y2​ 

∴ The locus of the foor of perpendicular is (x2 + y2)2 = 6x2 + 2y2​.

The correct answer is Option 3.

Concept:

The roots of any quadratic equation ax² + bx + c = 0 are equal when D = 0 i.e., b² - 4ac = 0. 

Calculation:

We have, ax + by = c

⇒ y = (c - ax)/b      ----(1)

On substituting this value in x2 + y2 = 16, we get,

⇒ x2 + (c - ax)² / b2 = 16

⇒ x2 + (c² - 2cax + a²x²) / b2 = 16

⇒ b²x2 + c2 - 2cax + a2x2 = 16b²

⇒ (a² + b²)x² - 2cax + (c² - 16b²) = 0      ----(2)

As the line is a tangent, there will be a single solution for the set of equations. Hence, the roots will be equal for the quadratic equation (2).

For equal roots, D = 0

⇒ (-2ca)² - 4 (a2 + b2)(c2 - 16b2) = 0

⇒ c²a² = (a2 + b2)(c2 - 16b2)

⇒ c2a² = c2a² + b²c² - 16a²b² - 16b⁴

⇒ 16a2b2 + 16b⁴ = b2c²

⇒ 16a2 + 16b² = c²

⇒ 16(a2 + b2) = c²

Hence, If ax + by = c is tangent to the circle x2 + y2 = 16 then 16(a2 + b2) = c².

Concept:

Equation of tangent to ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) at any point (a cosϕ, b sinϕ) is \(\frac{x}{a} \cos ϕ+\frac{y}{b} \sin ϕ=1\)

Calculation:

Given equation of curve is 9x² + 16y² = 144

⇒ \(\frac{x^2}{16}+\frac{y^2}{9}=1\)

The general point is (4 cos ϕ, 3 sin ϕ)

∴ The equation of tangent is \(\frac{x}{4} \cos ϕ+\frac{y}{9} \sin ϕ=1\)

∴ Coordinates of A = (4 secϕ, 0) and Coordinates of B = (0, 3 cosecϕ)

∴ AB = \(\sqrt{16 \sec ^2 ϕ+9 \operatorname{cosec}^2 ϕ}\) 

⇒ AB = \(\sqrt{16\left(1+\tan ^2 ϕ\right)+9\left(1+\cot ^2 ϕ\right)}\)

⇒ AB = \(\sqrt{25+(4 \tan ϕ)^2+(3 \cot ϕ)^2}\)

⇒ AB = \(\sqrt{25+(4 \tan ϕ-3 \cot ϕ)^2+24}\)

⇒ (AB)_min = \(\sqrt{25+24}\) = 7

∴ The minimum length of the line segment AB is 7.

Calculation:

The equation of the given ellipse is  4x2 + 5y2 = 20

It can be written as \(\frac{x^2}{5}+\frac{y^2}{4}=1\), which is comparable with the first standard form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

Here a² = 5 and b² = 4.

The given line is 3x + 2y - 5 = 0, its slope = -(3/2)

Since the tangents are perpendicular to the given line, the slope of the required tangents is 2/3 (from m₁ m₂ = -1)

Thus the equation of the required tangents is,

⇒ \(y = \frac{2}{3}x±\sqrt{5.(\frac{2}{3})^2+4}\)

⇒ \(y= \frac{2}{3}x±\sqrt{\frac{20}{9}+4}\)

⇒ \(y=\frac{2}{3}x±\frac{\sqrt{56}}{3}\)

⇒ \(3y=2x±\sqrt{56}\)

Additional Information Tangents to an Ellipse:

→ If the line y = mx + c touches the ellipse \(\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), then c2 = a2m2 + b2.

→ The straight lines \(\rm y = mx \pm \sqrt{a^2m^2 + b^2}\) represent the tangents to the ellipse.

Concept;

Equation of normal passes at \((ct_1, \frac{c}{t_1})\) is yt - xt³ - c + ct⁴  = 0

If it passes through any point say P having its coordinates as 't' i.e., \((ct_2, \frac{c}{t_2})\).

Calculation:

The coordinates must satisfy the equation. 

Hence, we get \(\frac{c}{t_2}\)t₁ - ct₂t₁³ - c + ct₁⁴  = 0

t₁ - t₁³t₂² - t₂ + t₂t₁⁴  = 0

Factorize this equation, (t₁ - t₂) + t₂t₁³(t₁ - t₂) = 0

On factorizing, we get, (t₁ - t₂)(t₂t₁³ + 1) = 0

Hence, t₂t₁³ + 1 = 0  ⇒ t₂t₁³  = -1 

or t₁ ≠ t₂ ⇒ This is not possible.

Hence,if the normals at the point \((ct_1, \frac{c}{t_1})\) on the hyperbola xy = c2 meets it again at the point \((ct_2, \frac{c}{t_2})\) then- t₂t₁³ = -1