Equation of a Tangent MCQ Quiz - Objective Question with Answer for Equation of a Tangent - Download Free PDF

Calculation

x² + y² – 2x + 4y – 4 = 0 

Centre (1, –2), r = 3

Reflection of (1, –2) about 2x – 3y + 5 = 0

\(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{-2(2+6+5)}{13}=-2\)

x = –3, y = 4

Equation of circle ‘C’

C : (x+3)² + (y– 4)² = 9

\(\ell(\operatorname{arcAB})=\frac{1}{6} \times 2 \pi \mathrm{r}\)

⇒ \(\mathrm{r} \theta=\frac{1}{6} \times 2 \pi \mathrm{r}\)

⇒ \(\theta=\frac{\pi}{3}\)

(α + 6)² + (β – 4)² = 27

⇒ \(\frac{(\alpha+3)^{2} \pm(\beta-4)^{2}=9}{(\alpha+6)^{2}-(\alpha+3)^{2}=18}\) 

⇒ 6α = –9 

⇒ \(\alpha=\frac{-3}{2}, \beta=\left(4-\frac{3 \sqrt{3}}{2}\right)\)

∴ \(\beta-\sqrt{3} \alpha\)

\(\left(4-\frac{3 \sqrt{3}}{2}\right)+\frac{3 \sqrt{3}}{2}\)

= 4

Hence option 4 is correct

Concept:

Common Tangent to a Hyperbola and a Parabola:

• The given problem asks for the equation of a common tangent to both the hyperbola and the parabola. To find this, we use the properties of the curves and the condition of tangency.
• The general equation of the hyperbola is x²/a² - y²/b² = 1, and for the parabola, it is y² = 8x. A common tangent means the line will touch both curves at exactly one point.
• The tangent to both curves with a positive slope can be derived by considering the slopes of the tangents to each curve and ensuring that they are equal at the point of tangency.

Calculation:

The equation of the hyperbola is:

x²/a² - y²/b² = 1

The equation of the parabola is:

y² = 8x

We are given that the tangent has a positive slope.

Let the equation of the common tangent be in the form y = mx + c, where m is the slope and c is the y-intercept.

For the parabola y² = 8x, the equation of the tangent with slope m is:

y = mx + c (tangent to the parabola)

For the hyperbola, we find the equation of the tangent similarly, ensuring that it satisfies the condition for tangency for both curves.

After solving the system of equations for the tangency condition, we find that the equation of the common tangent with a positive slope is:

y - 2x - 1 = 0

Hence, the correct answer is:

(2) y - 2x - 1 = 0

Calculation

\(\therefore CA = CB\)

\(\Rightarrow \frac{5}{\sqrt{10}} = \left|\frac{2m+1}{\sqrt{1+m^2}}\right|\)

squaring

\(\frac{25}{10} = \frac{4m^2+4m+1}{1+m^2}\)

\(\Rightarrow 5 + 5m^2 = 8m^2 + 8m + 2 \Rightarrow 3m^2 + 8m - 3 = 0 \Rightarrow m = \frac{1}{3}, -3\)

Equation of tangent OB is \(y = \frac{x}{3}\)

Hence option 3 is correct

Concept:

If the tangent is parallel to x-axis the slope is 0.

Solution:

Given the circle x2 + y2 - 2x - 4y + 1 = 0 .....(1)

Differentiating the above with respect to x

⇒ 2x + 2yy' - 2 - 4y' = 0

⇒ y'(y - 2) = 1 - x

⇒ \(y'=\frac{1-x}{y-2}\)

If the tangent is parallel to x-axis the slope is 0.

⇒ \(y'=\frac{1-x}{y-2}=0\)

⇒ \(\frac{1-x}{y-2}=0\)

⇒ x = 1

Put the above value of x in Eq. (1) to get the corresponding values of y

Thus,

1² + y² - 2(1) - 4y + 1 = 0

⇒ 1 + y² - 2 - 4y + 1 = 0

⇒ y² - 4y = 0

⇒ y(y - 4 ) = 0

⇒ y = 0, 4

Thus the two points are,

(1, 0) and (1, 4)

∴ The correct option is (3)

Concept

The distance of point (h,k) from the line ax + by + c = 0 is \( {|ah+bk+c| \over \sqrt{a^2 +b^2}}\)

The equation of a circle with centre (h,k) and radius R is,

 (x - h)2 + (y - k)2 = R2

Calculation:

The distance of centre (2, -1) from the tangent 3x + y = 0 is the radius of the circle,

⇒ R = \( {|3(2) +1(-1)| \over \sqrt{3^2 +1^2}}\)= \(\sqrt{5 \over 2}\)

Now the equation of the circle with centre (2,-1) and radius \(\sqrt{5 \over 2}\) is

(x - 2)2 + (y + 1)2 = \(5 \over 2\)  __(1)

Now let the equation of another tangent through origin be y = mx

Put in (1),

⇒ (x - 2)2 + (mx + 1)2 = \(5 \over 2\) 

⇒ (1 + m²)x2 + (2m - 4)x + 5 = \(5 \over 2\) 

⇒ (1 + m2)x2 + (2m - 4)x + \(5 \over 2\) = 0 which has only one solution.

⇒ Discriminant = (2m - 4)² - 4(1 + m2)\(5 \over 2\) = 0

⇒ 4m² - 16m + 16 - 10 - 10m² = 0

⇒ 6m2 + 16m - 6 = 0

⇒ 2(3m - 1)(m + 3) = 0

⇒ m = -3, \(1 \over 3\)

⇒ The equation of tangent other than 3x + y = 0 is x - 3y = 0

∴ The correct answer is option (3).

Concept:

The slope of the tangent to a curve y = f(x) is m = \(\rm dy\over dx\)

The slope of the normal = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)

Calculation:

Given curve y2 - 3x3 + 2 = 0

Differentiating the equation wrt x we get

⇒ 2y\(\rm dy\over dx\) - 9x2 + 0 = 0

⇒ 2y \(\rm dy\over dx\)  = 9x2

Slope at (1, -1)

⇒ 2(-1) \(\rm dy\over dx\) = 9 (1)

⇒ -2\(\rm dy\over dx\) = 9

⇒ \(\rm dy\over dx\) = -4.5

The slope of the tangent (m) = \(\rm {dy\over dx}\)

∴ m = -4.5

Concept:

Steps to find the equation of the tangent to the curve:

Find the first derivative of f(x).

Use the point-slope formula to find the equation for the tangent line.

Point-slope is the general form: y - y₁=m(x - x₁), Where m = slope of tangent = \(\rm \frac {dy}{dx}\) 

Calculation:

Given curve x3 + y2 + 3y + x = 0

Differentiating w.r.t x

3x² + 2y\(\rm dy\over dx\) + 3\(\rm dy\over dx\) + 1 = 0

(2y + 3)\(\rm dy\over dx\) = -3x² - 1

\(\rm dy\over dx\) = \(\rm -{3x^2+1\over2y+3}\)

At (2, -1)

\(\rm dy\over dx\) = \(\rm -{3(2)^2+1\over2(-1)+3}\)

\(\rm dy\over dx\) = \(-{12+1\over 1}\) = -13

The equation of the tangent is

(y - (-1)) = -13(x - 2)

y + 1 = -13x + 26

y + 13x - 25 = 0

Concept:

Tangent to a Circle: The equation of the tangent to a circle x² + y² = r² at a point (a, b) is given by ax + by = r².

Calculation:

Comparing the equation x2 + y2 = 16 with the general equation of the circle i.e. x2 + y2 = r², we have r² = 16.

Also, since the line ax + by = r2 is tangent to x2 + y2 = r2 at the point (a, b) on the circle, we can say that for the line ax + by = c to be a tangent to x2 + y2 = 16, (a, b) must be on the circle and c = r² = 16.

∴ a² + b² = c

⇒ c(a2 + b2) = c²

⇒ 16(a2 + b2) = c².

Concept:

For any straight line y= mx + c, m = slope or gradient and c = intercept

The equation of the tangent at the point (x1,y1) on the circle x2 + y2 = a² is given by:

xx1 + yy1 = a2

Calculation:

The given equation is,

x2 + y2 = a2

We know that for any given circle, the equation of the tangent at the point (x1,y1) on the circle is

xx1 + yy1 = a2

So, the given point is (h, h)

Hence, the equation of the tangent is,

hx + hy =a²

⇒y = -x + \(\rm a^2 \over h\)

Comparing the equation with y = mx + c the slope will be -1.

Concept:

Equation of the tangent at the point (x1, y1) to the curve x2 + y2 = a² is given by:

xx1 + yy1 = a2

Calculation:

The given equation is,

x2 + y2 = a²

We know that for any given circle, the equation of the tangent passing through the point (x₁, y₁) is

xx₁ + yy₁ = a²

So, the given point is (h,h)

Hence, the equation of the tangent is,

hx + hy =a2

Concept:

1. Tangent is perpendicular to the radius at the point of contact.

2. Angle between two lines: The angle θ between the lines having slope m1 and m2 is given by

\(\tan \theta = \;\left| {\frac{{{m_2} - {m_1}}}{{1 + \;{m_1}\;{m_2}}}} \right|\)

3. Sum of interior angles of a quadrilateral = 360° 

Calculation:

Given that, tangent to the circle at points A and B is represented by pair of straight lines PA & PB

x2 - 3y2 - 2x + 1 = 0 -----(1)

From equation (1)

x2 - 3y2 - 2x + 1 = 0

⇒ (x - 1)² = 3y²

⇒ x - 1 = ± √3y

Therefore, eqation of tangent PA and PB are

x + √3y - 1 = 0    ----(2)

x - √3y - 1 = 0    ----(3)

We know that, slope of line ax + by + c = 0 is -a/b. Therefore,

The slope of line (2) and (3) are -1/√3 and 1/√3 

Therefore, angle b/w tangents PA & PB

\(\theta \ =\ tan^{-1}[\frac{\frac{1}{\sqrt 3}\ -\ (\frac{-1}{\sqrt 3})}{1\ +\ (\frac{1}{\sqrt 3})(\frac{-1}{\sqrt 3})}]\)

\(⇒ \theta \ =\ tan^{-1}(\sqrt 3)\ =\ 60^∘\)

We know that,

∠ O + ∠ A + ∠ P + ∠ B = 360^∘ 

⇒ ∠ O = 360∘ - 180∘ - 60∘ 

⇒ ∠ O = 120∘

Hence, option 4 is correct.

Concept:

• The equation of a line, with slope m, is: y = mx + c.

• The distance between a point P(x₁, y₁) and the line ax + by + c = 0 is given by: Distance = \(\rm \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\).

• The equation of a circle with center at O(a, b) and radius r, is given by: (x - a)² + (y - b)² = r².
• Tangent to a Hyperbola: If the line y = mx + c touches the hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), then c² = a²m² - b². The equation of the tangent is: \(\rm y = mx \pm \sqrt{a^2m^2 - b^2}\). Either of the lines is the equation of the tangent but not both.

Calculation:

The equation of the circle can be written as (x - 4)2 + y2 = 42.

Comparing with the general form of a circle, we have center O(4, 0) and radius r = 4.

The equation of the given hyperbola can be written as \(\rm \frac{x^2}{3^2}-\frac{y^2}{2^2}=1\).

Comparing with the general form of a hyperbola, we have a = 3 and b = 2.

The equation of the tangent to this hyperbola will have the form:

\(\rm y = mx \pm \sqrt{a^2m^2 - b^2}\)

⇒ \(\rm y = mx \pm \sqrt{9m^2 - 4}\)

Since this line is a tangent to the circle as well, we must have:

Distance from the center O(4, 0) of the circle to the tangent \(\rm y = mx \pm \sqrt{9m^2 - 4}\) = radius (r = 4) of the circle.

Using the formula for the distance of a point from a line, we get:

\(\rm \frac{\left|m(4)+(-1)(0)\pm\sqrt{9m^2-4}\right|}{\sqrt{m^2+(-1)^2}}=4\)

⇒ \(\rm \left|4m\pm\sqrt{9m^2-4}\right|=4\sqrt{m^2+1}\)

On squaring both sides, we get:

⇒ \(\rm 16m^2 \pm 8m\sqrt{9m^2-4}+9m^2-4=16m^2+16\)

⇒ \(\rm \pm 8m\sqrt{9m^2-4}=20-9m^2\)

Squaring again, we get:

⇒ \(\rm 64m^2(9m^2-4)=400+81m^4-360m^2\)

⇒ \(\rm 576m^4-256m^2=400+81m^4-360m^2\)

⇒ \(\rm 495 m^4 + 104 m^2 - 400=0\)

⇒ \(\rm m^2 = \frac{-104 \pm \sqrt{104^2-4(495)(-400)}}{2\times495}\)

⇒ \(\rm m^2 = \frac{-104 \pm \sqrt{802816}}{990}\)

⇒ \(\rm m^2 = \frac{-104 \pm896}{990}\)

Discarding the negative value of m²:

⇒ \(\rm m^2 = \frac{-104 +896}{990}=\frac45\)

Since the slope is given to be positive, we get:

⇒ \(\rm m=\frac{2}{\sqrt5}\)

∴ Equation of the tangent will be:

\(\rm y = mx \pm \sqrt{9m^2 - 4}\)

⇒ \(\rm y = \frac{2}{\sqrt5}x \pm \sqrt{9\left(\frac45\right) - 4}\)

⇒ \(\rm y = \frac{2}{\sqrt5}x \pm\frac{4}{\sqrt5}\)

⇒ \(\rm 2x-\sqrt5y\pm4=0\).

Additional Information

• The slope (m) of the tangent at a point P(a, b) to a curve y = f(x), is given by: \(\rm m=\left(\frac{dy}{dx}\right)_{x=a,y=b}\).
• Tangent to a Parabola: The equation of the tangent to the parabola y2 = 4ax, at a point (x1, y1), is given by: yy1 = 2a(x + x1).

  Normal to a Parabola: The equation of the normal to the parabola y2 = 4ax, at a point (x1, y1), is given by: 2a(y - y1) = (-y1)(x - x1).

• Tangent to a Circle: The equation of the tangent to the circle x² + y² = r² at a point (x₁, y₁), is given by: xx₁ + yy₁ = r².

  Normal to a Circle: The equation of a normal to the circle x² + y² = r² at a point (x₁, y₁), is given by: yx₁ - xy₁ = 0.

• Tangent to an Ellipse: The equation of the tangent to the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1\).

  Normal to an Ellipse: The equation of the normal to the ellipse \(\rm \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1\).

• Tangent to a Hyperbola: The equation of the tangent to the hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1\).

  Normal to a Hyperbola: The equation of the normal to the hyperbola \(\rm \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), at a point (x₁, y₁), is given by: \(\rm \frac{a^2x}{x_1}+\frac{b^2y}{y_1}= a^2+b^2\).

Concept:

The slope of the tangent to a curve y = f(x) is m = \(\rm dy\over dx\)

The slope of the normal = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)

Calculation:

Given curve y2 - 3x3 + 2 = 0

Differentiating the equation wrt x we get

⇒ 2y\(\rm dy\over dx\) - 9x2 + 0 = 0

⇒ 2y \(\rm dy\over dx\)  = 9x2

Slope at (1, -1)

⇒ 2(-1) \(\rm dy\over dx\) = 9 (1)

⇒ -2\(\rm dy\over dx\) = 9

⇒ \(\rm dy\over dx\) = -4.5

The slope of the tangent (m) = \(\rm {dy\over dx}\)

∴ m = -4.5

Concept:

Steps to find the equation of the tangent to the curve:

Find the first derivative of f(x).

Use the point-slope formula to find the equation for the tangent line.

Point-slope is the general form: y - y₁=m(x - x₁), Where m = slope of tangent = \(\rm \frac {dy}{dx}\) 

Calculation:

Given curve x3 + y2 + 3y + x = 0

Differentiating w.r.t x

3x² + 2y\(\rm dy\over dx\) + 3\(\rm dy\over dx\) + 1 = 0

(2y + 3)\(\rm dy\over dx\) = -3x² - 1

\(\rm dy\over dx\) = \(\rm -{3x^2+1\over2y+3}\)

At (2, -1)

\(\rm dy\over dx\) = \(\rm -{3(2)^2+1\over2(-1)+3}\)

\(\rm dy\over dx\) = \(-{12+1\over 1}\) = -13

The equation of the tangent is

(y - (-1)) = -13(x - 2)

y + 1 = -13x + 26

y + 13x - 25 = 0

Concept:

Tangent to a Circle: The equation of the tangent to a circle x² + y² = r² at a point (a, b) is given by ax + by = r².

Calculation:

Comparing the equation x2 + y2 = 16 with the general equation of the circle i.e. x2 + y2 = r², we have r² = 16.

Also, since the line ax + by = r2 is tangent to x2 + y2 = r2 at the point (a, b) on the circle, we can say that for the line ax + by = c to be a tangent to x2 + y2 = 16, (a, b) must be on the circle and c = r² = 16.

∴ a² + b² = c

⇒ c(a2 + b2) = c²

⇒ 16(a2 + b2) = c².