Locus MCQ Quiz in मराठी - Objective Question with Answer for Locus - मोफत PDF डाउनलोड करा

Let us consider a point \(P(h,k)\) in the first quadrant.

such that \(d(P,A)=d(P,O)\)

Therefore

\(|h-3|+|k-2|=h+k\) ...(i)

Case 1:

If \(h < 3, k < 2\), then \((h, k)\) lies in region l. Then,

\(3-h+2-k=h+k\)

\(h+k=\dfrac{5}{2}\)

Case 2:

If \(h > 3, k < 2\), then \((h, k)\) lies in region II. Then,

\(h-3+2-k=h+k\)

\(k=\dfrac{-1}{2}\) ...(this is not possible since P is in the first quadrant)

Case 3:

If \(h > 3, k > 2\) then \((h, k)\) lies in region III. Then,

\(h+k-5=h+k\) ...(hence no solution)

Case 4:

If \(h < 3, k > 2\), then \((h, k)\) lies in region IV. Then,

\(3-h+k-2=h+k\)

\(h=\dfrac{1}{2}\)

Replacing \((h,k)\) by \((x,y)\) we get

\(x+y=\dfrac{5}{2}\) ...(of finite length, part which is in the first quadrant)

and \(x=\dfrac{1}{2}\)

Thus the locus of point P is the union of these two.

Then by the given conditions,

\((h-a_1)^2 + (k-b_1)^2 = (h-a_2)^2 + (k-b_2)^2\)

\(\Rightarrow 2 h (a_1 - a_2)+2k(b_1 - b_2) + a_2^2 -a_{1}^{2}+ b_2^2 - b_1^2 =0\)

\(\Rightarrow h (a_1 -a_2) + k (b_1 -b_2) + \dfrac{1}{2} (a_2^2 +b_2^2 - a_1^2 - b_1^2)=0\)

Also, since \((h, k)\) lies on the given locus, therefore

\((a_1 - a_2)x + (b_1 - b_2)y +c = 0\)

Comparing Eqs. (i) and (ii), we get

\(c = \dfrac{1}{2} (a_2^2 + b_2^2 -a_1^2 - b_1^2)\)

\(x \cos \alpha + y \sin \alpha = P\)

Let \(h = \dfrac{P}{2 \cos \alpha}\), \(k = \dfrac{P}{2 \sin \alpha}\)

\(\cos \alpha = \dfrac{P}{2h}\), \(\sin \alpha = \dfrac{P}{2k}\)

\(\cos^2 \alpha = \dfrac{P^2}{4h^2}\), \(\sin^2 \alpha = \dfrac{P^2}{4k^2}\)

\(\cos^2 \alpha + \sin^2 \alpha = \dfrac{P^2}{4h^2} + \dfrac{P^2}{4k^2}\)

\(\Rightarrow \dfrac{P^2}{4x^2} + \dfrac{P^2}{4y^2} = 1\) or \(\dfrac{1}{x^2} + \dfrac{1}{y^2} = \dfrac{4}{P^2}\)

Calculation

sin²y = cos²x

siny = ± cosx

If sin y = cos x = sin(\(\frac{\pi}{2}\) - x)

⇒ y = nπ + (-1)ⁿ(\(\frac{\pi}{2}\) - x)

If sin y = -cos x = sin(x - \(\frac{\pi}{2}\))

⇒ y = nπ + (-1)ⁿ(x - \(\frac{\pi}{2}\))

Hence option 3 is correct

Given:

The ratio of distances of point P(x, y) from point (1, 1) and line x − y + 2 = 0 is 1 : √2

Concept:

Use distance formulas to relate point-to-point and point-to-line distances.

Formula Used:

Distance between two points: √[(x₂ − x₁)² + (y₂ − y₁)²]

Distance from point to line ax + by + c = 0: |ax + by + c| / √(a² + b²)

Calculation:

Let P(x, y) be the variable point

Distance from P to (1, 1): √[(x − 1)² + (y − 1)²]

Distance from P to line x − y + 2 = 0: |x − y + 2| / √2

Given ratio = 1 : √2

⇒ √[(x − 1)² + (y − 1)²] = (1 / √2) × |x − y + 2|

⇒ √[(x − 1)² + (y − 1)²] = |x − y + 2| / √2

⇒ Square both sides:

⇒ (x − 1)² + (y − 1)² = (x − y + 2)² / 2

⇒ x² − 2x + 1 + y² − 2y + 1 = (x² − 2xy + y² + 4x − 4y + 4) / 2

⇒ x² + y² − 2x − 2y + 2 = (x² − 2xy + y² + 4x − 4y + 4) / 2

⇒ Multiply both sides by 2:

⇒ 2x² + 2y² − 4x − 4y + 4 = x² − 2xy + y² + 4x − 4y + 4

⇒ Move RHS to LHS:

⇒ (2x² − x²) + (2y² − y²) + (−4x − 4x) + (−4y + 4y) + 2xy + (4 − 4)

⇒ 3x² + 3y² − 8x + 2xy = 0

∴ The equation of the locus is:

3x² + 2xy + 3y² − 12x − 4y + 4 = 0

Calculation

Equation of LM:

y - 2 =\(\frac{ t }{ 2}\) (x - t)

⇒ M(0, \(\frac{4 - t^2 }{ 2}\))

Midpoint of LM: N(h, k)

⇒ 2h = t, 2k = 4 - \(\frac{t^2 }{ 2}\)

⇒ x² = 2 - y

Hence option 2 is correct

Explanation:

∴ The locus of point P will be a straight line which is the right bisector of AB.

Concept used: 

Angle subtended by the diameter of a circle at a point on the circumference is Right angle (90°).

.

Explanation:

Join AB and draw a circle with AB as diameter. The locus of point P is the circumference of this circle, since Angle subtended by the diameter of a circle at a point on the circumference is Right angle (90°). Therefore, whatever be the position of point P, the measure of ∠ APB will always be 90°.

∴ The locus of point P is the circumference of the circle with AB as diameter.

Concept:

The distance of a point (h,k) from a line Ax + By + C = 0 is \(d = \left |{ Ah+Bk + C \over \sqrt{A^2 + B^2}}\right|\)

The distance between two point (x₁,y₁) and (x₂,y₂) is \(\sqrt {(x_2-x_1)^2 +(y_2 -y_1)^2}\)

Calculation:

Let the moving point be (x,y)

then its distance from (4,0) is

\(\sqrt {(x-4)^2 +y^2}\)

and the distance of the point (x,y) from the line x = 16

that is, the distance of the point from the line x + 0y - 16 = 0 is 

\(d = \left |{ x - 16\over \sqrt{(1)^2 + (0)^2}}\right|\)

⇒ \(d = \left |{ x - 16}\right|\)

Given that, the distance from the point (4, 0) is half that of its distance from the line x = 16

⇒ \(\sqrt {(x-4)^2 +y^2} = {1 \over 2}|{x -16 }|\)

Squaring both sides,

⇒ \({(x-4)^2 +y^2} = {1 \over 4}|{x -16 }|^2\)

⇒ 4(x² - 8x + 16 + y²) = x² - 32x + 256

⇒ 3x2 + 4y2 = 192

∴ The correct option is (1).

Concept:

Locus: The locus is the collection of all those points whose position is defined by a particular condition.

Internal sectional formula: Let P (x₁, y₁) and Q (x₂, y₂) be the endpoints of the given line segment PQ and R(x, y) be the point which divides PQ in the ratio m: n.

Then, the coordinates of R (x, y) are \(\left\{\frac{m × x_{2} + n × x_{1}}{m + n}, \frac{m × y_{2} + n × y_{1}}{m + n}\right\}\).

Explanation:

Let R(x, y) be the mid-point of OM and (α, β ) be the co-ordinate of the point M. Then

x =\(\frac{0+α}{2}\) and y =\(\frac{0+β}{2}\) 

⇒ α = 2x  and β = 2y

Therefore, the coordinates of M are (2x, 2y).

The slope of OM = \(\frac{2y-0}{2x-0}=\frac{y}{x}\)

and Slope of MP =\(\frac{2y-b}{2x-a}\)

Since OM is perpendicular to MP. Therefore

Slope of OM × Slope of MP = -1

⇒ \(\frac{y}{x} \times \frac{2y-b}{2x-a}=-1\)

⇒ y(2y - b) =  - x (2x - a)

⇒ 2y² - by = - 2x² + ax

⇒ 2x² + 2y² - ax - by = 0

Therefore, the locus of the midpoint of OM is 2x2 + 2y2 - ax - by = 0