Question

find the equation of locus of a point which is equidistant from the point A( and B (0 4​

Answer 1

Answer: Let P(h,k) be any point in the locus.

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we have

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2 =0.

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2 =0.Thus, locus of P is x

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2 =0.Thus, locus of P is x 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2 =0.Thus, locus of P is x 2 −y

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2 =0.Thus, locus of P is x 2 −y 2

Let P(h,k) be any point in the locus.Let PM be the perpendicular distance of P fro x-axis, i.e.,PM=∣x∣Let PNeft be the perpendicular distance of P fro y-axis, i.e.,PN=∣y∣∵PM=PN(Given)⇒∣x∣=∣y∣Squaring both sides, we havex 2 =y 2 ⇒x 2 −y 2 =0.Thus, locus of P is x 2 −y 2 =0.

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